on monomial burnside rings · primitive idempotents of cb(a,g) under these maps. in the paper [9]...

ON MONOMIAL BURNSIDE RINGS

a thesis

submitted to the department of mathematics

and the institute of engineering and science

of bilkent university

in partial fulfillment of the requirements

for the degree of

master of science

By

Ergun Yaraneri

September, 2003

I certify that I have read this thesis and that in my opinion it is fully adequate,

in scope and in quality, as a thesis for the degree of Master of Science.

Assoc. Prof. Dr. Laurance J. Barker (Supervisor)



Asst. Prof. Dr. Ergun Yalcın



Asst. Prof. Dr. Semra Kaptanoglu

Approved for the Institute of Engineering and Science:

Prof. Dr. Mehmet B. BarayDirector of the Institute Engineering and Science

ii

ABSTRACT

ON MONOMIAL BURNSIDE RINGS

Ergun Yaraneri

M.S. in Mathematics

Supervisor: Assoc. Prof. Dr. Laurance J. Barker

September, 2003

This thesis is concerned with some different aspects of the monomial Burnside

rings, including an extensive, self contained introduction of the A−fibred G−sets,

and the monomial Burnside rings. However, this work has two main subjects that

are studied in chapters 6 and 7.

There are certain important maps studied by Yoshida in [16] which are very

helpful in understanding the structure of the Burnside rings and their unit groups.

In chapter 6, we extend these maps to the monomial Burnside rings and find the

images of the primitive idempotents of the monomial Burnside C−algebras. For

two of these maps, the images of the primitive idempotents appear for the first

time in this work.

In chapter 7, developing a line of research persued by Dress [9], Boltje [6],

Barker [1], we study the prime ideals of monomial Burnside rings, and the prim-

itive idempotents of monomial Burnside algebras. The new results include;

(a): If A is a π−group, then the primitive idempotents of Z(π)B(A,G) and

Z(π)B(G) are the same

(b): If G is a π′−group, then the primitive idempotents of Z(π)B(A,G) and

QB(A,G) are the same

(c): If G is a nilpotent group, then there is a bijection between the primitive

idempotents of Z(π)B(A,G) and the primitive idempotents of QB(A,K) where

K is the unique Hall π′−subgroup of G.

(Z(π) = {a/b ∈ Q : b /∈ ∪p∈πpZ}, π =a set of prime numbers).

Keywords: Monomial Burnside rings, ghost ring, primitive idempotents, inflation

map, invariance map, orbit map, conjugation map, restriction map, induction

map, prime ideals, prime spectrum .

iii

OZET

TEK TERIMLI BURNSIDE HALKALARI

Ergun Yaraneri

Matematik, Yuksek Lisans

Tez Yoneticisi: Doc. Dr. Laurance J. Barker

Eylul, 2003

Bu tezde tek terimli Burnside halkalarının degisik yonlerini inceledik. Fakat bu

calısma iki onemli konu icermektedir, ve bunlar 6. ve 7. kısımlarda ele alınmıstır.

Burnside halkaları uzerinde onemli fonksiyonlar tanımlanmıstır. 6. kısımda

bu fonksiyonları Burnside halkalarını alt halka olarak iceren tek terimli Burnside

halkalarına genislettik. Ayrıca yine 6. kısımda tek terimli Burnside C− cebir-

lerinin ilkel idempotentlerinin genislettigimiz fonksiyonlar altındaki goruntulerini

bulduk. Soz konusu fonksiyonlardan ikisi icin ilkel idempotentlerin goruntuleri

ilk olarak bu calısmada yer almaktadır.

Kısım 7 de ise tek terimli Burnside halkalarının asal ideallerini inceledik

ve bazı tek terimli Burnside cebirlerinin ilkel idempotentleri hakkında bilgiler

edindik. Elde ettigimiz sonuclar daha onceden baska calısmalarda yer almayan

yeni sonuclar da icermektedir. 7. kısımdaki bu yeni sonuclar arasında asagıdaki

uc sonuc en onemlileridir. (Z(π) = {a/b ∈ Q : b /∈ ∪p∈πpZ}, π =asal sayılardan

olusan bir kume).

(a): Eger A bir π−grup ise Z(π)B(A,G) ve Z(π)B(G) aynı ilkel idempotentlere

sahiptir.

(b): Eger G bir π′−grup ise Z(π)B(A,G) ve QB(A,G) aynı ilkel idempotentlere

sahiptir

(c): Eger G bir nilpotent grup ise Z(π)B(A,G) ve QB(A,K) ’nin ilkel idempo-

tentleri arasinda bire-bir esleme yapabiliriz. Burada K G’nin biricik Hall π′−alt

grubudur.

Anahtar sozcukler : Tek terimli Burnside Halkaları, hayalet halka, ilkel idem-

potentler, infilasyon foksiyonu, stabil elemanlar foksiyonu, yorunge fonksiyonu,

eslenik foksiyonu, daraltma fonksiyonu, genisletme fonksiyonu, asal idealler, asal

idealler spekturumu.

iv

Acknowledgement

I would like to thank my supervisor Laurance J. Barker, firstly for introducing

me to this area of algebra, and for all his patience and advice thereafter. His

influence can be felt in some chapters of this work.

My thanks are also due to Semra Kaptanoglu and Ergun Yalcın who accepted

to read this thesis.

I would like to thank Mustafa Kesir and Olcay Coskun for their help with

typing difficulties.

v

Contents

1 Introduction 2

2 A−Fibred G−Sets 6

3 Monomial Burnside Rings 12

4 Possible Ghost Rings 18

5 Primitive Idempotents of CB(A,G) 28

6 Some Maps 30

6.1 The Inflation Map . . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.2 The Invariance Map . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.3 The Orbit Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6.4 The Conjugation Map . . . . . . . . . . . . . . . . . . . . . . . . 42

6.5 The Restriction Map . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.6 The Induction Map . . . . . . . . . . . . . . . . . . . . . . . . . . 44

vi

CONTENTS 1

7 Prime Ideals Of B(A,G) 45

8 Some Further Maps 83

8.1 B(G)→ B(A,G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

8.2 B(A,G)→ B(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

8.3 B(A,G)→ B(A′, G) . . . . . . . . . . . . . . . . . . . . . . . . . 85

8.4 B(A,H)→ B(A,G) . . . . . . . . . . . . . . . . . . . . . . . . . 87

8.5 B(A,G)→ B(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

8.6 B(A,G)→ B(A,AG) . . . . . . . . . . . . . . . . . . . . . . . . . 90

8.7 B(A,G)→ B(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

8.8 B(A,G)→ B(AG) . . . . . . . . . . . . . . . . . . . . . . . . . . 92

8.9 B(A1 × A2, G)→ B(A1, G)×B(A2, G) . . . . . . . . . . . . . . . 92

8.10 The Number Of Orbits . . . . . . . . . . . . . . . . . . . . . . . . 93

9 The Ring B(A,G) 95

Chapter 1

Introduction

The concept of fibred permutation sets arises naturally in topics closely connected

with many aspects of representation theory: character theory, induction theorems

etc.

The theory of fibred permutation sets for a finite group G is a quite easy

extension of the theory of permutation sets where the role played by points of the

permutation sets is now played by fibres, which are copies of a fixed finite abelian

group A called the fibre group. We define an A−fibred G−set to be a finite

A−free A × G−set. We concentrate exclusively on the study of isomorphism

classes of A−fibred G−sets. These classes may be added and multiplied in a

natural fashion and in this manner they generate a commutative ring known as

the monomial Burnside ring of A and G, and denoted by B(A,G). In the special

case where A is trivial, we recover the ordinary Burnside ring B(1, G) = B(G).

The ordinary Burnside rings have many uses in representation theory, the the-

ory of G−spheres, and group theory. They are Mackey functors. The importance

of the ordinary Burnside rings in such areas of algebra leads to an extension of

the ordinary Burnside ring. The first extension was given by Dress in [9].

Following Dress [9], the monomial Burnside rings, explicitly or implicitly, have

been studied in contexts related to induction theorems. See, for instance, Boltje

2

CHAPTER 1. INTRODUCTION 3

[2], [3], [4], [5], [6].

The main contributor to the subject is undoubtedly Dress who introduced the

monomial Burnside rings, and discovered a number of deep and striking results

in [9]. One of his celebrated results in [9] asserts that a finite group G is solvable

if and only if the monomial Burnside ring B(A,G) has no nontrivial idempotents.

Because of the importance of idempotents, the idempotents of the ordinary

Burnside rings have received much attentions, first by Dress in [8], a formula

for the primitive idempotents of QB(G) in terms of the transitive basis of B(G)

appeared for the first time in [10] by Gluck. After that, in [15], Yoshida found a

formula for the primitive idempotents of RB(G) where R is any integral domain

of characteristic 0 from which the idempotent formula of Gluck follows as an easy

corollary. For the monomial Burnside rings there is a similar history. In [6], Boltje

gave an idempotent formula for the primitive idempotents of KB(A,G) in terms

of the transitive basis ofKB(A,G) whereK is a field of characteristic 0 containing

enough roots of unity, and A is the unit group of an algebraically closed field.

Later, in [1], Barker gave idempotent formulas for the primitive idempotents of

CB(A,G), KB(A,G) and B(A,G) in terms of the transitive basis of B(A,G)

where K is any field of characteristic 0 from which the idempotent formula of

Boltje follows as an easy consequence. Since B(1, G) = B(G), the idempotent

formula of QB(G) obtained by Gluck in [10] follows from the idempotent formula

of Barker in [1].

The monomial Burnside rings introduced by Dress in [9] are more general than

the monomial Burnside rings considered by Boltje, Barker, and us. We consider

the same monomial Burnside rings as Barker [1].

We study some different aspects of the monomial Burnside rings and try to

extend some theory from the Burnside rings. We made much use of the paper

[9] by Dress, which is a fundamental paper on this subject especially in chapters

2, 3, 4 and 7. However because of the full generalities of Drees’ paper [9], these

chapters, while influenced by [9], have different flavors.

Chapter 2 contains an extensive account of the theory of fibred permutation


sets.

In chapter 3 and 4, the monomial Burnside ring is defined and its basic prop-

erties are studied.

There are certain maps defined on the Burnside rings which appear in [16].

With these maps the Burnside rings become Mackey functors. In chapter 6, we

extend these maps to the monomial Burnside rings and find the images of the

primitive idempotents of CB(A,G) under these maps.

In the paper [9] by Dress, there is a section dealing with prime ideals of his

ring. Because of the full generalities of his ring and his ghost ring, in chapter 7

our approach is indeed different from his approach, although inspired from his

paper. Although the name of chapter 7 is prime ideals, our main object there is

finding the primitive idempotents of the monomial Burnside rings tensored over

Z with an integral domain of characteristic 0. We give some partial answers when

the integral domain satisfies some restrictive conditions.

In chapter 8, we give some maps whose domains or codomains are the mono-

mial Burnside rings.

In chapter 9, we give some ring theoretic propertis of the monomial Burnside

rings.

Finally, let us summarize some of the new results in this thesis. In chapter

1, we prove all the results about A−fibred G−sets whose proofs are left to the

reader in Dress [9] and Barker [1]. In chapter 2, and 3, we try to exhibit which

results in [1] can be deduced from [9]. In chapter 6, we extend the maps studied

by Yoshida in [16] to the monomial Burnside rings and find the images of the

primitive idempotents of the monomial Burnside C−algebras. For two of these

maps, the images of the primitive idempotents appear for the first time in this

work. The results obtained in chapter 7 include new facts which did not appear

in the papers [9] by Dress, and [1] by Barker. The new results obtained in

chapter 7 includes, for instance, some facts about the primitive idempotents of

Z(π)B(A,G) where A is a π−group, or G is a π′−group, or G is a nilpotent group


(Here, Z(π) = {a/b ∈ Q : b /∈ ∪p∈πpZ}, π =a set of prime numbers).

To facilitate the reading, important definitions and results have been repeated

where necessary.

Chapter 2

A−Fibred G−Sets

The monomial Burnside rings were introduced by Dress in [9]. We will see in the

next chapter that the elements of the monomial Burnside rings are closely related

to the A−fibred G−sets. So we first need to introduce an account of the theory

of A−fibred G−sets. In [9], Dress gave a very short introduction to A−fibred

G−sets leaving details to the reader. Also he considered more general A−fibred

G−sets than we want to consider here. However, we mainly follow [9] but using

the notations of Barker in [1].

In this chapter we introduce A−fibred G−sets and give some properties. We

need some facts about G−sets. The following facts about G−sets are well-known

and can be found in [14]. Let G be a finite group.

(1) Let G be a group. A finite set S is called a G−set if there is a map G×S → S,

(g, s) 7→ gs, satisfying; 1s = s and (gh)s = g(hs) for all g, h ∈ G, s ∈ S.(2) Let S and T be G−sets. A map f : S → T is called a G−map if f(gs) = gf(s)

for all g ∈ G, s ∈ S.(3) Let S be G−set. For any s ∈ S, we write orbG(s) = {gs : g ∈ G} and

stabG(s) = {g ∈ G : gs = s}. They are called G−orbit of s and G−stabilizer of

s, respectively. Moreover, orbG(s) is a G−set and stabG(s) is a subgroup of G.

(4) A G−set S is said to be transitive if for any s1, s2 ∈ S there is a g ∈ G such

that gs1 = s2.

(5) A G−set S is transitive if and only if any G−map from a G−set T into S is

6

CHAPTER 2. A−FIBRED G−SETS 7

surjective.

(6) For any subgroup H of G, the set of left cosets of H in G becomes a G−set

by left multiplication.

(7) Let S be a G−set. For any s ∈ S, the map G/stabG(s) → orbG(s) given

by gstabG(s) 7→ gs is a bijective G−map (G−isomorphism) and so the G−sets

G/stabG(s) and orbG(s) are isomorphic. Hence in particular, |G : stabG(s)| =

|orbG(s)|. We write T1 'G T2 for isomorphic G−sets T1 and T2.

(8) Let S and T be G−sets. Take any s ∈ S and t ∈ T. Then orbG(s) 'G orbG(t)

if and only if stabG(s) =G stabG(t).

(9) For a G−set S and a subgroup H of G we write SH to denote the set of

H−fixed points of S.

(10) For G−sets S and T, HomG(S, T ) denotes the set of all G−maps from S to

T.

(11) Let S be a G−set and H be a subgroup of G. Then we have a bijection

between the sets HomG(G/H, S) and SH given by (f : G/H → S) 7→ f(H).

(12) For H,K ≤ G; (G/K)H = {gK : H ≤ gK} = {gitjK : 1 = 1, ..., r; j =

1, ..., s} where g1Kg−11 , ..., grKg

−1r are all distinct G−conjugates of K containing

H and t1, ..., ts are the left coset representatives of K in NG(K).

(13) Let H,K be subgroups of G. Then (G/K)H =⊎H≤W,W=GK

(G/K)W .

(14) LetH1, ..., Hn be all distinct nonconjugate subgroups ofG, and S be aG−set.

Then S '⊎i λi(G/Hi) where λi = |Si|

|G:Hi| and Si = {s ∈ S : stabG(s) =G Hi}.(15) (Burnside)For any G−set S, the number of G−orbits of S is 1

|G|∑

g∈G |S<g>|.(16) A G−set S is called G−free if stabG(s) = 1 for all s ∈ S. For such G−sets

each G−orbits have the same number of elements.

(17) Let S be a G−set. Writing S as a disjoint union of its G−orbits we can

express S in the form GX = {gx : g ∈ G, x ∈ X} where X is a set of orbit

representatives. Once an X is chosen, any element of S can be written uniquely

in the form gx where g ∈ G, x ∈ X.

Now we can begin to study A−fibred G−sets. Let A be a finite abelian group

and G be a finite group. We write AG for A × G by identifying a ∈ A with

(a, 1) ∈ A×G and g ∈ G by (1, g) ∈ G. Note that by our notational convention


ag = ga for any a ∈ A and g ∈ G. A finite A−free AG−set is called an A−fibred

G−set, and its A−orbits orbA(s), s ∈ S, are called fibres. We sometimes use the

notation As instead of orbA(s) for fibres. Let S be an A−fibred G−set. Writing

S as a disjoint union of its fibres we can express S as AX = {ax : a ∈ A, x ∈ X}where X is a set of representatives of fibres. Any element of S = AX can be

written uniquely in the form ax where a ∈ A, x ∈ X. Hence, given any A−fibred

G−set S we can see S as a set of formal products AX = {ax : a ∈ A, x ∈ X},and a1x1 = a2x2 if and only if a1 = a2, x1 = x2. Consequently, by an A−fibred

G−set (equivalently by an A−free AG−set) we mean a set of formal products

AX = {ax : a ∈ A, x ∈ X} such that AX is a G−set and X is a finite set.

Let AX be an A−fibred G−set. Since it is an AG−set, it must be isomorphic to

a disjoint union of the sets of left cosets of some subgroups of AG. However, it

is not true for all subgroups of AG that the set of left cosets forms an A− fibred

G−set because it may not be A−free.

Remark 2.1 Let V ≤ G and ν ∈ Hom(V,A). Then {ν(v−1)v : v ∈ V } is a

subgroup of AG, and the set of its left cosets in AG forms an A−fibred G−set.

Proof : Put 4(V,ν) = {ν(v−1)v : v ∈ V }. It is clear that 4(V,ν) is a subgroup

of AG and AG/4(V,ν) is an AG− set (by left multiplication). Hence we only

need to check its A−stabilizers. Take any ν(v−1)v ∈ 4(V,ν). Then a ∈ A is in

stabA(ν(v−1)v) if and only if aν(v−1)v = ν(v−1)v. Thus, stabA(ν(v−1)v) = 1 and

so AG/4(V,ν) is an A−fibred G−set.

�

We use the notation 4(V,ν) for the subgroup {ν(v−1)v : v ∈ V } of AG for any

V ≤ G and ν ∈ Hom(V,A). Later we will show that any transitive A−fibred

G−set is AG−isomorphic to AG/4(V,ν) for some V ≤ G and ν ∈ Hom(V,A).

For an A−fibred G−set AX, the set of its A−orbits (fibres) {Ax : x ∈ X} is

denoted by A \ AX.

Remark 2.2 Let AX be an A−fibred G−set. Then A \AX is a G−set with the

action;

G× (A \ AX)→ (A \ AX), (g, Ax) 7→ gAx = Agx.


Proof : Obvious.

�

Note that for an A−fibred G−set AX we have the following immediate proper-

ties;

orbAG(ax) = orbAG(x), stabAG(ax) = stabAG(x), orbG(Aax) = orbG(Ax), and

stabG(Aax) = stabG(Ax) for all a ∈ A, x ∈ X.

Remark 2.3 Let AX be an A−fibred G−set. Then AX is a transitive AG−set

if and only if A \ AX is a transitive G−set.

Proof : (⇒) Take any two fibres Ax1 and Ax2. Since AX is a transitive AG−set

and x1, x2 ∈ AX there is an ag ∈ AG such that agx1 = x2. But then gAx1 = Ax2

and so A \ AX is a transitive G−set.

(⇐) Take any a1x1, a2x2 ∈ AX. Since A \ AX is a transitive G−set, there is a

g ∈ G such that gAx1 = Ax2. But then a2x2 = ga3x1 for some a3 ∈ A and so

(a−11 a3g)(a1x1) = a2x2 implying that AX is a transitive AG−set.

�

We call AX a transitive A−fibred G−set if AX is a transitive AG−set, or

equivalently if A \ AX is a transitive G−set.

Remark 2.4 Let S = AX be an A−fibred G−set, and s ∈ S. Then;

(i) The map πs : stabAG(s)→ stabG(As) given by πs(ag) = g is a group isomor-

phism,

(ii) For any g ∈ stabG(As), there is a unique ag ∈ A such that aggs = s,

(iii) The map νs : stabG(As) → A given by νs(g) = a−1g is a group homomor-

phism,

(iv) stabAG(s) = 4(stabG(As),νs),

(v) AG/4(stabG(As),νs) is a transitive A−fibred G−set,

(vi) AG/4(stabG(As),νs) 'AG orbAG(s).

Proof : (i) It is a straightforward checking.

(ii) Since πs is bijective, for any g ∈ stabG(As) there is a unique ag ∈ A such that


πs(agg) = g. Note that aggs = s.

(iii) Given g, h ∈ stabG(As), we find unique elements ag, ah, agh ∈ A such that

aggs = ahhs = agh(gh)s = s by part (ii). Then s = agh(gh)s = aghg(hs) =

agha−1h (gs) = agha

−1h a−1

g s and so agh = agah because S is A−free. Hence νs ∈Hom(stabG(As), A).

(iv) If ag ∈ stabAG(s), then g ∈ stabG(As) and s = ags = aggs = νs(g−1)gs.

Thus as = νs(g−1)s, and since S is A−free νs(g

−1) = a. So ag = νs(g−1)g ∈

{νs(g−1)g : g ∈ stabG(As)} = 4(stabG(As),νs). Therefore, stabAG(s) is contained in

the set 4(stabG(As),νs). Converse direction is clear because an element νs(g−1)g of

4(stabG(As),νs) is equal to agg where aggs = s.

(v) Clearly it is a transitive AG−set. Moreover it is A−free from 2.1.

(vi) Obvious.

�

For any A−fibred G−set S we know from 2.4 that the A−fibred G−sets orbAG(s)

and AG/4(stabG(As),νs) are isomorphic (as AG−sets) where νs is the uniquely de-

termined element of Hom(stabG(As), νs) by the condition: gs = νs(g)s for all

g ∈ stabG(As). Hence, in particular any transitive A−fibred G−set S is isomor-

phic to AG/4(stabG(As),νs) where s is any element of S. And conversely for any

V ≤ G and ν ∈ Hom(V,A) the set AG/4(V,ν) is a transitive A−fibred G−set.

We use the notation AνG/V to denote the transitive A−fibred G−set AG/4(V,ν),

and use [AνG/V ] to denote its isomorphism class. Also for any A−fibred G−set

AX, we write [AX] for the isomorphism class of AX.

Remark 2.5 Let S = AX be an A−fibred G−set, and for s ∈ S let νs :

stabG(As) → A be the uniquely determined element of Hom(stabG(As), A) by

the condition: gs = νs(g)s for all g ∈ stabG(As). Then

νgs = gνs and νas = νs for any g ∈ G and a ∈ A.

Proof : h ∈ stabG(As) if and only if Ahs = As, or equivalently A(ghg−1)gs =

Ags. Hence h ∈ stabG(As) if and only if ghg−1 ∈ stabG(Ags). Moreover, hs =

νs(h)s and (ghg−1)gs = νgs(ghg−1)gs imply that νs(h)s = νgs(ghg

−1)s. Since S

is A−free, νgs(ghg−1) = νs(h) implying that νgs = gνs.

Note that A(as) = As. So the functions νas and νs have the same domain. Also


g(as) = νas(g)(as) implies that gs = νas(g)s. On the other hand, gs = νs(g)s and

so νas(g)s = νs(g)s. Because S is A−free, νas = νs.

�

Let ch(A,G) = {(V, ν) : V ≤ G, ν ∈ Hom(V,A)}. Then G acts on ch(A,G)

by conjugation; (g, (V, ν)) 7→ g(V, ν) = (gV, gν) where gν : gV → A is given

by gν(gvg−1) = ν(v) for all v ∈ V. We write (V, ν) =G (W,ω) if the elements

(V, ν), (W,ω) ∈ ch(A,G) are in the same G−orbit of ch(A,G).

Remark 2.6 Let (V, ν), (W,ω) ∈ ch(A,G). Then,

AνG/V 'AG AωG/W (equivalently, [AνG/V ] = [AωG/W ]) if and only if

(V, ν) =G (W,ω).

Proof : AνG/V 'AG AωG/W if and only if the subgroups 4(V,ν),4(W,ω) of AG

are AG−conjugates. Now if ah4(V,ν) = 4(W,ω) for some ah ∈ AG, then

{ν(g−1)hgh−1 : g ∈ V } = {ω(g−1)g : g ∈ W}.But {ν(g−1)hgh−1 : g ∈ V } = {hν((hgh−1)−1)(hgh−1) : g ∈ V } = {hν(u−1)u :

u ∈ hV }. So, hV = W and hν = ω implying that h(V, ν) = (W,ω).

�

Consider AνG/V which denotes the A−fibred G−set AG/4(V,ν) where 4(V,ν)

= {ν(v−1)v : v ∈ V }. Let us denote ag4(V,ν) by ag4. Two fibres A(ag4) and

A(bh4) are equal if and only if there is a c ∈ A such that cg4 = h4 if and only

if h−1g ∈ V and c = ν(g−1h) ∈ A. Thus, the fibres A(ag4) and A(bh4) are equal

if and only if gV = hV. Hence, we have a bijective map A \ (AνG/V ) → G/V

given by A(ag4) 7→ gV. It is clear that this map is a G−map. Consequently,

A \ (AνG/V ) 'G G/V.Suppose AνG/V is given. It can be written in the form AX where X is a set of

A−orbits representatives. Since it is A−free, each A−orbit has the same number

of elements which is |A|, and we showed above that the number of A−orbits is

equal to |G/V |. Therefore, to represent the A−fibred G−set AνG/V = AG/4(V,ν)

in the form AX we can take for exampleX as the set {g4(V,ν) : gV ⊆ G} (AG acts

on AX by left multiplication). Note that stabG(A(g4(V,ν))) = gV, νg4(V,ν)= gν

and |AνG/V | = |A||G : V |.

Chapter 3

Monomial Burnside Rings

We are still assuming that A is a finite abelian group and G is a finite group. We

introduce two binary operations on A−fibred G−sets. The most obvious one is

the disjoint union. The other one is slightly more complicated than the disjoint

union, and we introduce it now as exactly Dress did in [9] but we are using the

notations of [1].

Suppose S = AX and T = AY are both A−fibred G−sets. Then their

cartesian product S × T is an A−set with the action: a(s, t) = (as, a−1t).

Let S⊗T denote the set of A−orbits of the cartesian product S×T with respect

to the above A−action. We write s⊗ t for the A−orbit containing (s, t) ∈ S×T.Thus,

S ⊗ T = {s⊗ t : s ∈ S, t ∈ T}, s⊗ t = {(as, a−1t) : a ∈ A}.

Note that (as)⊗ t = s⊗ (at) for any a ∈ A and (s, t) ∈ S × T.We let AG act on S ⊗ T as: ag(s⊗ t) = (ags)⊗ (gt).

Remark 3.1 S ⊗ T constructed above is an A−fibred G−set.

Proof : The action is well-defined: Let s ⊗ t = s′ ⊗ t′ for some s, s′ ∈ S and

t, t′ ∈ T. We want to show that ag(s ⊗ t) = ag(s′ ⊗ t′) for any ag ∈ AG. Since

12

CHAPTER 3. MONOMIAL BURNSIDE RINGS 13

s ⊗ t = s′ ⊗ t′, (s, t) and (s′, t′) are in the same A−orbit of S × T. Thus there

is an a ∈ A such that (s′, t′) = (a−1s, at). But then (ags′, gt′) = (gs, agt) and so

ags′ ⊗ gt′ = gs⊗ agt = ags⊗ gt. Hence ag(s⊗ t) = ag(s′ ⊗ t′).S ⊗ T is A−free: Take an element s⊗ t ∈ S ⊗ T and compute its A−stabilizer.

a ∈ A is in the A−stabilizer of s⊗t if and only if as⊗t = s⊗t which is to say that

(as, t), (s, t) ∈ S×T are in the same A−orbit of S×T. Then, (as, t), (s, t) ∈ S×Tare in the same A−orbit of S×T if and only if (bas, b−1t) = (s, t) for some b ∈ A,or equivalently a = b = 1. Hence, S ⊗ T is A−free.

The action properties are satisfied: It is obvious that 1(s ⊗ t) = s ⊗ t and

((ag)(bh))s⊗ t = (ag)((bh)s⊗ t).�

Theorem 3.2 For any (V, ν), (W,ω) ∈ ch(A,G) we have;

AνG/V ⊗ AωG/W 'AG⊎

V gW⊆G

Aν.gωG/V ∩ gW.

Proof : Remember that AνG/V = AG/4(V,ν) and AωG/W = AG/4(W,ω). Put

4(V,ν) = 4, 4(V,ν) = 4′, and S = AνG/V ⊗ AωG/W. Since we can express any

A−fibred G−set S as a disjoint union of its AG−orbits, and since orbAG(s) 'AGAG/stabAG(s) = AG/4(stabG(As),νs); we can proceed as follows.

Take any element ag4 ⊗ bh4′ of S. Then orbAG(ag4 ⊗ bh4′) = orbAG(4 ⊗abg−1h4′). Hence we calculate stabAG(s) for elements s ∈ S of the form 4⊗g4′.

ah ∈ AG is in stabAG(4 ⊗ g4′) if and only if ah4 ⊗ hg4′ = 4 ⊗ g4′. But

ah4⊗ hg4′ = 4⊗ g4′ if and only if (ah4, hg4′) and (4, g4′) are in the same

A−orbit of AνG/V ×AωG/W which is equivalent to, (ah4, hg4′) = (b4, b−1g4′)

for some b ∈ A. Now (ah4, hg4′) = (b4, b−1g4′) is the same as with ab−1h ∈4 and bg−1hg ∈ 4′, or equivalently h ∈ V, ν(h−1) = ab−1, g−1hg ∈ W and

ω((g−1hg)−1) = b. Hence, ah ∈ AG is in stabAG(4⊗g4′) if and only if h ∈ V ∩gWand a = ν(h−1)ω((g−1hg)−1) = (ν.gω)(h−1). Therefore,

stabAG(4⊗ g4′) = 4(V ∩gW,ν.gω) and so orbAG(4⊗ g4′) 'AG Aν.gωG/V ∩ gW.

Now, orbAG(4⊗ g4′) = orbAG(4⊗h4′) if and only if there is an ak ∈ AG such

that ak4 ⊗ kg4′ = 4 ⊗ h4′, which is to say that (ak4, kg4′) and (4, h4′)

are in the same A−orbit of AνG/V × AωG/W. But this holds if and only if


there is a b ∈ A such that (ak4, kg4′) = (b4, b−1h4′), which is equivalent to

ab−1k ∈ 4 and bh−1kg ∈ 4′. Using the definitions of 4 and 4′, we see that

ab−1k ∈ 4 and bh−1kg ∈ 4′ if and only if k ∈ V, ν(k−1) = ab−1, h−1kg ∈ W and

ω((h−1kg)−1) = b, or equivalently h−1kg ∈ W and k ∈ V. But then by k ∈ V,

orbAG(4⊗ g4′) = orbAG(4⊗ h4′) if and only if V hW = V gW. Hence,

AνG/V ⊗ AωG/W 'AG⊎

V gW⊆G

Aν.gωG/V ∩ gW.

�

The formula in 3.2 is known as the Mackey product formula.

Define an addition and multiplication on the isomorphism classes of A−fibred

G−sets as follows:

[S] + [T ] = [S ] T ] and [S][T ] = [S ⊗ T ].

It is clear that the above operations are well-defined, commutative, associative,

and moreover the multiplication is distributive over the addition. Thus, the

set of isomorphism classes of A−fibred G−sets forms a commutative semiring.

We write B(A,G) for the associated Grothendieck ring and call it monomial

Burnside ring. Therefore B(A,G) is a set of formal differences of isomorphism

classes of A−fibred G−sets, and it is a commutative ring with 1 with respect to

the following operations;

[AX] + [AY ] = [A(X ] Y )] and [AX][AY ] = [AX ⊗ AY ] = [A(X ⊗ Y )]

where the action of AG on A(X ⊗ Y ) is given by ag(x⊗ y) = agx⊗ y.Note that the multiplicative identity of the ring is [AτG/G] where τ is the trivial

group homomorphism from G to A.

Remember that ch(A,G) = {(V, ν) : V ≤ G, ν ∈ Hom(V,A)} is a G−set by

conjugation.

Remark 3.3 (i) B(A,G) is a commutative ring with 1.

(ii) B(A,G) is a free Z−module with so called transitive basis {[AνG/V ] : (V, ν) ∈ch(A,G)}.(iii) B(A,G) =

⊕(V,ν)∈Gch(A,G) Z[AνG/V ] where the notation under the direct

sum means that (V, ν) runs over a set of representatives of nonconjugate elements

of ch(A,G).


(iv) The multiplication of B(A,G) on its transitive basis given as;

[AνG/V ][AωG/W ] =∑

V gW⊆G

[Aν.gωG/V ∩ gW ].

Proof : In chapter 2 we proved the following three facts.

(1) Any transitive A−fibred G−set is isomorphic to AνG/V for some (V, ν).

(2) Any set of the form AνG/V is a transitive A−fibred G−set.

(3) AνG/V ' AωG/W if and only if (V, ν) =G (W,ω).

Hence; (i), (ii), and (iii) follows from the above three and from the definition of

B(A,G). We proved (iv) in 3.2.

�

For a G−set S, let AS = {as : a ∈ A, s ∈ S} be the set of formal products.

Thus, a1s1 = a2s2 if and only if a1 = a2 and s1 = s2. We let AG act on AS

as: (bg)(as) = (ab)(gs) for all bg ∈ AG and as ∈ AS. Then, AS becomes an

A−fibred G−set. If [S] denotes the isomorphism class of the G−set S, then it is

clear that [S] = [T ] implies [AS] = [AT ]. So we have a well-defined map

ψ1 : B(G)→ B(A,G) given by ψ1([S]) = [AS] for any G−set S.

Remark 3.4 (i) ψ1([G/V ]) = [AτG/V ] for any V ≤ G, where τ is the trivial

element of the group Hom(V,A).

(ii) ψ1 is a unital ring monomorphism.

(iii) B(G) can be regarded as a subring of B(A,G).

Proof : (i) ψ1([G/V ]) = [A(G/V )], A(G/V ) = {a(gV ) : a ∈ A, gV ∈ G/V },and AτG/V = AG/4(V,τ) where 4(V,τ) = {1v : v ∈ V } ≤ AG.

Define a map f : A(G/V ) → AG/4(V,τ) where f(a(gV )) = ag4(V,τ). The ele-

ments a(gV ) and b(hV ) of A(G/V ) are equal if and only if a = b and gV = hV

which is to say that (bh)−1(ag) = 1(h−1g) ∈ 4(V,τ). Thus, the elements a(gV )

and b(hV ) of A(G/V ) are equal if and only if ag4(V,τ) = bh4(V,τ). Hence, f is

well-defined, and injective.

Obviously from its definition, f is surjective and an AG−map.


Now since we proved that A(G/V ) and AG/4(V,τ) are isomorphic (by the map

f) we have ψ1([G/V ]) = [AτG/V ].

(ii) It follows easily from the multiplication formula given in 3.3 (iv).

(iii) Since ψ1 is a unital ring monomorphism, by identifying B(G) with ψ1(B(G))

we can regard B(G) as a subring of B(A,G).

�

Remember that for any A−fibred G−set AX, the set A \AX of its A−orbits

(fibres) is a G−set with respect to the G−action: gAx = Agx. Also in chapter 2

we showed that A \ (AνG/V ) 'G G/V. Hence we have a well-defined map

ψ2 : B(A,G)→ B(G) given by ψ2([AX]) = [A \ AX].

Remark 3.5 (i) ψ2([AνG/V ]) = [G/V ] for any (V, ν) ∈ ch(A,G).

(ii) ψ2 is a unital ring epimorphism.

Proof : Follows immediately from the above explanation and from the multipli-

cation formula given in 3.3 (iv).

�

Let A′ and A be two finite abelian groups such that A′ ≤ A. Hence for any V ≤G, a group homomorphism ν : V → A′ can be seen as a group homomorphism

V → A. By this way the A′−fibred G−set A′νG/V can be seen as the A−fibred

G−set AνG/V. Moreover, if A′νG/V 'A′G A′

ωG/W then (V, ν) =G (W,ω) and

so AνG/V 'AG AωG/W. Thus we have a well-defined map ψ3 : B(A′, G) →B(A,G) given by ψ3([A

′νG/V ]) = [AνG/V ]. It is clear that ψ3 is a unital ring

monomorphism.

We constructed the ring homomorphisms ψ1 and ψ2 in 3.4 and 3.5. Now

we consider the composition map φ = ψ1 ◦ ψ2 : B(A,G) → B(A,G) where

φ([AνG/V ]) = [AτG/V ] for any (V, ν) ∈ ch(A,G) where τ denotes the trivial

homomorphism.


Remark 3.6 (i) φ is a unital ring homomorphism, and it can be also seen as

B(G)−module endomorphism of the B(G)−module B(A,G).

(ii) φ is a projection onto B(G) ≤ B(A,G).

(iii) B(A,G) = B(G)⊕Ker(φ) as B(G)−modules or Z−modules.

Proof : All parts are obvious. (Note that B(G) is a unital subring of B(A,G)

and Ker(φ) is an ideal of B(A,G). So, the decomposition in (iii) is not merely a

submodule decomposition. )

�

We close this chapter after giving an elementary consequence of the mul-

tiplication formula given in 3.3 (iv). For any ν ∈ Hom(G,A), we have

[AνG/G]n = [AνnG/G]. Since ν(g) ∈ A for any g ∈ G, [AνG/G]|A| = 1.

Therefore, [AνG/G] is a unit in B(A,G) for any ν ∈ Hom(G,A). Moreover

K = {[AνG/G] : ν ∈ Hom(G,A)} is a multiplicatively closed subset of B(A,G)

containing 1. Hence K is a subgroup of the unit group B(A,G)∗ of B(A,G). In

fact, this shows that Hom(G,A) embeds in B(A,G)∗ by ν 7→ [AνG/G] for any

ν ∈ ch(A,G).

Chapter 4

Possible Ghost Rings

The Burnside ring B(G) can be embedded in the ring Zn where n is the number of

noncojugate subgroups of G. That is, there is a ring monomorphism from B(G)

to Zn and so the image of B(G) is a subring of Zn and it is called the ghost

ring of B(G). That is why the name of this chapter is possible ghost rings. In

this chapter we embed B(A,G) to two rings which are easier to work with than

B(A,G). The first one is a direct product of some group rings, and the second

one which is easier is a direct product of C. The first one was studied in [9] and

the second one in [1]. We will mainly follow [9] for the first ghost ring using the

notations in [1] and supply details skipped in [9]. We are still assuming that A

is a finite abelian group. However, for the second ghost ring we have to assume

that A is cyclic. Since the monomial Burnside rings introduced by Dress in [9] are

more general than the monomial Burnside rings we are considering, the second

ghost ring introduced by Barker in [1] will be used more than the first ghost ring

in the next chapters.

For any H ≤ G, let ZHom(H,A) be the group ring. Conjugation by an ele-

ment g ∈ G induces a group isomorphism Hom(H,A)→ Hom(gH,A) which can

be extended to a group ring isomorphism from ZHom(H,A) to ZHom(gH,A).

For an A−fibred G−set S = AX and s ∈ S, remember that νs is the uniquely

determined element of Hom(stabG(As), A) by the condition: gs = νs(g)s for all

g ∈ stabG(As).

18

CHAPTER 4. POSSIBLE GHOST RINGS 19

Lemma 4.1 Let AX and AY be A−fibred G−sets. For any x ∈ X and y ∈ Ywe have;

(i) stabG(Ax⊗ y) = stabG(Ax) ∩ stabG(Ay),

(ii) νx⊗y = νxνy.

Proof : (i) Let g ∈ G. If gx = a1x and gy = a2y for some a1, a2 ∈ A, then

g(x⊗y) = a1a2(x⊗y) and, more generally, ga(x⊗y) = aa1a2(x⊗y) for all a ∈ A.On the other hand, if g(x⊗ y) = b(x⊗ y) for some b ∈ A, then gx = cx for some

c ∈ A, and we must have gy = c−1by. So we have shown that g(Ax⊗ y) = Ax⊗ yif and only if gAx = Ax and gAy = Ay. So part (i) follows.

(ii) For any g ∈ stabG(Ax ⊗ y) = stabG(Ax) ∩ stabG(Ay), we have gx = νx(g)x,

gy = νy(g)y and g(x ⊗ y) = νx⊗y(g)(x ⊗ y). Then νx⊗y(g)(x ⊗ y) = g(x ⊗ y) =

(gx)⊗ (gy) = (νx(g)x)⊗ (νy(g)y) = νx(g)νy(g)(x⊗y). Since AX⊗AY is A−free,

it follows that νx⊗y(g) = νx(g)νy(g).

�

For any H ≤ G, we define a map from B(A,G) to ZHom(H,A) as:

ψH : B(A,G)→ ZHom(H,A), [AX] 7→∑

x∈X,H≤stabG(Ax)

νx|H

where νx|H denotes the restriction of νx to H. We usually omit |H and use νx for

νx|H .

Lemma 4.2 ψH is well-defined.

Proof : Suppose [AX] = [AY ]. We want to show that ψ([AX]) = ψ([AY ]). Now,

AX and AY are two isomorphic AG−sets. Thus, there is a bijective AG−map

f : AX → AY. For any x ∈ X, there are unique elements ax ∈ A and yx ∈ Y

such that f(x) = axyx.

Note that for x, x′ ∈ X if yx = yx′ then f(ax′x) = ax′axyx and f(axx′) = axax′yx′ .

So f(ax′x) = f(axx′) implying that x = x′ and ax = ax′ . That is, Y = {yx : x ∈

X} and yx are all distinct where x range in X.

Take an x ∈ X. Then g ∈ stabG(Ax) if and only if Agx = Ax which is to say


that Agf(x) = Af(x) (equivalently g ∈ stabG(Af(x))), because f respects the

AG−action. Thus the maps νx and νf(x) are defined in the same domain, and

stabG(Ax) = stabG(Af(x)) = stabG(Ayx). Moreover, for any g ∈ stabG(Ax) we

have gx = νx(g)x and gf(x) = νf(x)(g)f(x). On the other hand from gx = νx(g)x

we get f(gx) = f(νx(g)x) implying that gf(x) = νx(g)f(x).Hence, νf(x)(g)f(x) =

νx(g)f(x). Since AY is A−free, νf(x)(g) = νx(g). So we proved that νx = νf(x) =

νaxyx = νyx (the last equality follows from 2.5).

Finally, using Y = {yx : x ∈ X}, stabG(Ax) = stabG(Ayx), and νx = νyx we

compute

ψ([AX]) =∑

x∈X,H≤stabG(Ax)

νx =∑

x∈X,H≤stabG(Ayx)

νyx

=∑

y∈Y,H≤stabG(Ay)

νy = ψ([AY ]).

�

Theorem 4.3 (i) ψH is a unital ring homomorphism.

(ii) For any (V, ν) ∈ ch(A,G);

ψH([AνG/V ]) =∑

gV⊆G,H≤gV

gν.

(iii) If H �G V, then ψH([AνG/V ]) = 0.

Proof : (i) Additivity is clear because [AX] + [AY ] = [A(X ] Y )].

Let AX and AY be A−fibred G−sets. Then using 4.1;

ψH([AX][AY ]) = ψH([AX ⊗ AY ]) = ψH([A(X ⊗ Y )])

=∑

x⊗y∈X⊗Y,H≤stabG(Ax⊗y)

νx⊗y

=∑

x∈X,y∈Y,H≤stabG(Ax),H≤stabG(Ay)

νxνy

= (∑

x∈X,H≤stabG(Ax)

νx)(∑

y∈Y,H≤stabG(Ay)

νy)

= ψH([AX])ψH([AY ]).


(ii) Remember that AνG/V = AG/4(V,ν) where 4(V,ν) = {ν(v−1)v : v ∈ V }.In chapter 2 we showed that to write AνG/V in the form AX we can take

X = {g4(V,ν) : gV ⊆ G}. Now we easily find that stabG(A(g4(V,ν))) = gV

and νg4(V,ν)= gν. So, ψH([AνG/V ]) =

∑gV⊆G,H≤gV

gν.

(iii) It is obvious because {gV ⊆ G : H ≤ gV } is empty set if H �G V.

�

Since conjugation by an element g ∈ G induce a group ring isomorphism

ZHom(H,A) → ZHom(gH,A), conjugation by g changes the map ψH :

B(A,G) → ZHom(H,A) to a map, gψH : B(A,G) → ZHom(gH,A), by tak-

ing g−conjugates of the image of ψH . Note that gψH = ψgH because:

ψgH([AνG/V ]) =∑

kV⊆G,gH≤kV

kν =∑

g−1kV⊆G,H≤g−1kV

g(g−1kν)

= g(∑

g−1kV⊆G,H≤g−1kV

g−1kν) = gψH([AνG/V ]).

Note that for any H ≤ G, the group NG(H)/H = N(H) acts on Hom(H,A) by

conjugation: (gH, ν) 7→ gν : gH → A where gν(gh) = ν(h) for all h ∈ H. Thus

by Z−linear extension, N(H) acts on ZHom(H,A).

Remark 4.4 The action of N(H) on ZHom(H,A) fixes ψH(B(A,G)) setwise.

Proof : From the explanation above we have gψH = ψgH for any g ∈ G. Hence

the result follows because gH = H for gH ∈ N(H).

�

Remark 4.5 (i) ψH([AνG/H]) =∑

gH⊆NG(H)gν for any ν ∈ Hom(H,A).

(ii) ψH([AνG/H]) = |stabN(H)(ν)|∑

ω ω where ω ranges over all distinct

N(H)−conjugates of ν.

(iii) For any (V, ν) ∈ ch(A,G),

ψH([AνG/V ]) =∑

H≤W≤G,W=GV

ψW ([AνG/V ]).


Proof : (i) Obvious.

(ii) It is clear because Hom(H,A) is an N(H)−set by conjugation.

(iii) ψH([AνG/V ]) =∑

gV⊆G,H≤gVgν. Note that the indices of the sum range

in the set {gV ⊆ G : H ≤ gV } = (G/V )H . We can write (G/V )H as

]H≤W,W=GV (G/V )W (this property is stated in the beginning of chapter 2, (13)).

So,

{gV ⊆ G : H ≤ gV } = ]H≤W,W=GV {gV ⊆ G : W ≤ gV }. Then

ψH([AνG/V ]) =∑

gV⊆G,H≤gV

gν

=∑

H≤W,W=GV

(∑

gV⊆G,W≤gV

gν)

=∑

H≤W≤G,W=GV

ψW ([AνG/V ]).

�

Now we show that the product map∏

H≤G ψH is an injective ring homomorphism

from B(A,G) to∏

H≤G ZHom(H,A). We need the following lemma.

Let F be a subset of the subgroups of G such that if V ∈ F , then gH ∈ F for

any H ≤ V, and g ∈ G. We put

B(A,G,F) =⊕

(V,ν)∈Gch(A,G),V ∈F

Z[AνG/V ].

Lemma 4.6

B(A,G,F) =⋂H/∈F

Ker(ψH).

Proof : If H �G V then ψH([AνG/V ]) = 0. Hence,

B(A,G,F) ⊆⋂H/∈F

Ker(ψH).

Take any z =∑

(W,ω)∈Gch(A,G) λW,ω[AωG/W ] ∈ B(A,G) such that z is in⋂H/∈F Ker(ψH) where λW,ω is an integer for (W,ω) ∈ ch(A,G). We want to show

that z ∈ B(A,G,F). So it suffices to show that if λW,ω 6= 0, then W ∈ F . Assume

the contrary. Then there is a W /∈ F with λW,ω 6= 0 for some ω ∈ Hom(W,A).


Let V0 be such an element which is maximal with respect to ≤G . Thus for some

ν0 ∈ Hom(V0, A) λV0,ν0 6= 0, V0 /∈ F , and if λV,ν 6= 0 and V /∈ F then V0 �G V.

Now,

0 = ψV0(z) =∑

(W,ω)∈Gch(A,G)

λW,ωψV0([AωG/W ]).

Because H �G V implies that ψH([AνG/V ]) = 0,

0 = ψV0(z) =∑

(W,ω),V0≤GW


For any W appearing in the last sum, if W ∈ F then V0 ∈ F which is not the

case. So,

0 = ψV0(z) =∑

(W,ω),V0≤GW,W /∈F


By the maximality of V0,

0 = ψV0(z) =∑

ω∈N(V0)Hom(V0,A)

λV0,ωψV0([AωG/V0]).

Then using 4.5(ii);

0 = ψV0(z) =∑

ω∈N(V0)Hom(V0,A)

λV0,ω|stabN(V0)(ω)|∑µ

µ

where for a fixed ω ∈ Hom(V0, A) the index µ ranges over all distinct

N(V0)−conjugates of ω. So all µ appearing in the last sum are distinct. Since the

elements of Hom(V0, A) are linearly independent over Z, we must have λV0,ν0 = 0

which is a contradiction. Hence we proved that

B(A,G,F) ⊇⋂H/∈F

Ker(ψH).

�

Theorem 4.7 The map∏

H≤G ψH : B(A,G)→∏

H≤G ZHom(H,A) is an injec-

tive ring homomorphism.

Proof : Let F be the empty set. Then 4.6 implies that

0 = B(A,G,F) =⋂H≤G

Ker(ψH) = Ker(∏H≤G

ψH).


So the product map is injective.

�

Let R be a commutative ring with 1, we write RB(A,G) for R⊗Z B(A,G) and

RHom(H,A) for R ⊗Z ZHom(H,A). We denote the R−linear extension of the

map ψH by again ψH . The results 4.6 and 4.7 are still true when Z is replaced

by any commutative ring R with identity such that |G| is not a zero divisor of R

(proofs are exactly the same).

For any (V, ν) ∈ ch(A,G) put nV,ν = |stabN(V )(ν)|. Note that for (V, ν) =G (W,ω)

nV,ν = nW,ω. Now for any H ≤ G and (V, ν) ∈ ch(A,G), by 4.5(iii);

ψH([AνG/V ]) =∑

H≤W≤G,W=GV

ψW ([AνG/V ]).

Suppose W1, ...,Wn are all distinct conjugates of V containing H. Let Wi = giV.

Then since gψH = ψgH ,

ψH([AνG/V ]) =∑i

giψV ([AνG/V ]).

Using 4.5(i) and (ii);

ψH([AνG/V ]) =∑i

gi(∑

gV⊆NG(V )

gν) =∑i

gi(|stabN(V )(ν)|∑ω

ω).

Hence for any (V, ν) ∈ ch(A,G); ψH( 1nV,ν

[AνG/V ]) ∈ ZHom(H,A) and

ψV ( 1nV,ν

[AνG/V ]) = [ν]+ where [ν]+ denotes the sum of all distinct conjugates of

ν. Let ZHom(V,A)N(V ) be the set of N(V )−fixed points of ZHom(V,A). Then

obviously it is a subring of ZHom(V,A) and it is a free Z−module with basis

[ν]+. Hence, images of the different elements 1nV,ν

[AνG/V ] under the map ψV form

a Z−basis of ZHom(V,A)N(V ).

Remark 4.8 The Z−linear span of the set { 1nV,ν

[AνG/V ] : (V, ν) ∈ ch(A,G)} is

a subring of QB(A,G) and isomorphic to∏

V≤GGZHom(V,A)N(V ).

Proof : It follows from the explanation above. Note that the product is taken

over all nonconjugate subgroups of G. It is because: G acts on∏

V≤G ZHom(V,A)

and g sends the term ZHom(V,A) onto ZHom(gV,A) isomorphicly. Thus


since gψV = ψgV , the map∏

V≤G ψV : QB(A,G) →∏

V≤G QHom(V,A) maps

the Z−linear span of { 1nV,ν

[AνG/V ] : (V, ν) ∈ ch(A,G)} isomorphicly onto

(∏

V≤G ZHom(V,A))G . Also note that we have

(∏V≤G

ZHom(V,A))G '∏V≤GG

ZHom(V,A)N(V ).

�

Now we study the ghost ring introduced in [1]. For this purpose we have to

assume that A is cyclic. Hence, from now on in this chapter A is a finite cyclic

group and we assume that A ≤ C∗.

We have already some algebra maps ψH : CB(A,G) → CHom(H,A) and we

want to construct algebra maps from CB(A,G) to C. For any h ∈ H, de-

fine a map ev(h) : CHom(H,A) → C given by ev(h)(∑

ν∈Hom(H,A) λνν) =∑ν∈Hom(H,A) λνν(h). It is clear that ev(h) is a C−algebra epimorphism. Note

that ev(h1) = ev(h2) if and only if ν(h1) = ν(h2) for all ν in Hom(H,A)

if and only if h−12 h1 ∈ Kerν for all ν ∈ Hom(H,A). Let for any V ≤ G,

O(V ) = ∩ν∈Hom(V,A)Kerν. We saw above that ev(v1) = ev(v2) if and only if

v1O(V ) = v2O(V ).

For any H ≤ G and h ∈ H define a map SGH,h : CB(A,G) → C as SGH,h =

ev(h) ◦ ψH . It is clear that SGH,h is a C−algebra homomorphism (because it is a

composition of two such maps), and

SGH,h([AνG/V ]) = ev(h)(∑

gV⊆G,H≤gV

gν) =∑

gV⊆G,H≤gV

gν(h).

Remark 4.9 For any H ≤ G and h ∈ H we have;

(i) For any [AνG/V ] ∈ B(A,G)

SGH,h([AνG/V ]) =∑

gV⊆G,H≤gV

gν(h),

(ii) For any A−fibred G−set AX

SGH,h([AX]) =∑

x∈X,H≤stabG(Ax)

νx(h),


(iii) SGH,h : CB(A,G)→ C is a C−algebra epimorphism,

(iv) SGH,h = SGgH,gh for any g ∈ G.

Proof : (i) It is proved above.

(ii) Using the definition SGH,h = ev(h) ◦ ψH we can easily find the desired result

because we know the rules of the maps ev(h) and ψH .

(iii) Because it is a composition of two C−algebra maps, the result follows.

(iv) SGgH,gh = ev(gh) ◦ gψH = ev(gh) ◦ ψgH = SGH,h.

�

We define the following two sets (first one is already defined before);

ch(A,G) = {(V, ν) : V ≤ G, ν ∈ Hom(V,A)},

el(A,G) = {(H, h) : H ≤ G, hO(H) ∈ H/O(H)}

where O(H) = ∩ν∈Hom(H,A)Kerν, or equivalently O(H) is the minimal normal

subgroup of H such that H = H/O(H) is an abelian group of exponent dividing

|A|. The sets ch(A,G) and el(A,G) are called the set of A−subcharacters of G

and the set of A−subelements of G, respectively. See [1] for a more detailed

explanation of these two sets. We just state the following.

Remark 4.10 (i) The sets ch(A,G) and el(A,G) are G−sets by the conjugation

action of G.

(ii) |ch(A,G)| = |el(A,G)|, |G \ ch(A,G)| = |G \ el(A,G)| and |Hom(H,A)| =|H|, where G \ el(A,G) and G \ el(A,G) denote G−orbit representatives.

Proof : See [1].

�

We write (H, h) =G (K, k) if the elements (H, h), (K, k) ∈ el(A,G) are in the

same G−orbit of el(A,G).


Theorem 4.11 For any (H, h), consider the map SGH,h : KB(A,G) → K given

by SGH,h([AX]) =∑

x∈X,H≤stabG(Ax) νx(h) where K is a field of characteristic 0

containing enough roots of unity to ensure that A ≤ K∗. Then;

(i) For any (V, ν) ∈ ch(A,G),

SGH,h([AνG/V ]) =∑

gV⊆G,H≤gV

gν(h).

(ii) SGH,h is a K−algebra epimorphism.

(iii) Any K−algebra homomorphism from KB(A,G) to K is of the form SGH,h for

some (H, h) ∈ el(A,G).

(iv) SGH,h = SGK,k if and only if (H, h) =G (K, k).

(v) KB(A,G) is a semisimple algebra.

(vi) The following map is a C−algebra isomorphism,∏(H,h)∈Gel(A,G)

SGH,h : CB(A,G)→∏

(H,h)∈Gel(A,G)

C.

Proof : Indeed we know (i), (ii) and half of (iv) from 4.9. For the rest, or for all

of them see [1].

�

Theorem 4.11 which was obtained by Barker in [1] is very important, and it will

be used in the next chapters.

Chapter 5

Primitive Idempotents of

CB(A,G)

We are still assuming that A is a finite cyclic group and A ≤ C∗. An explicit

formula for the primitive idempotents of CB(A,G) in terms of the transitive

basis can be found in [1]. We just state in this chapter some results that we need

in later chapters, for details see [1].

From 4.11 (vi) we know that ϕ =∏

(H,h)∈Gel(A,G) SGH,h is a C−algebra iso-

morphism from CB(A,G) to∏

(H,h)∈Gel(A,G) C. We know that B = {[AνG/V ] :

(V, ν) ∈ ch(A,G)} is a C−basis of the C−algebra CB(A,G). Let B′ be the stan-

dard basis of∏

(H,h)∈Gel(A,G) C. That is, it consists of all vectors (0, .., 1, .., 0) with

only one nonzero entry which is 1 in the (H, h)th place. Suppose we order Band B′. Then ϕ has an n × n matrix say B[ϕ]B′ with respect to the ordered ba-

sis B and B′ where n = |G \ ch(A,G)| = |G \ el(A,G)|. Thus we have for any

z ∈ CB(A,G) that [ϕ(z)]B′ = B[ϕ]B′ [z]B where [ϕ(z)]B′ denotes the coordinate

matrix of ϕ(z) with respect to B′, and [z]B denotes the coordinate matrix of z

with respect to B. Since ϕ is an isomorphism, primitive idempotents maps onto

primitive idempotents. Also the primitive idempotents of∏

(H,h)∈Gel(A,G) C are

just the elements of B′. Hence ϕ−1(B′) must be the set of primitive idempotents

of CB(A,G). Hence, in concrete examples we can find the primitive idempotents

28

CHAPTER 5. PRIMITIVE IDEMPOTENTS OF CB(A,G) 29

of CB(A,G) by evaluating the inverse of the matrix B[ϕ]B′ .

Remark 5.1 (i) The primitive idempotents of CB(A,G) are of the form eGH,h

where (H, h) runs over all nonconjugate A−subelements of G.

(ii) eGH,heGK,k =

{1 if (H, h) =G (K, k)

0 otherwise.

(iii)∑

(H,h)∈Gel(A,G) eGH,h = 1.

(iv) eGH,h is the unique element of CB(A,G) satisfying the following condition for

all (K, k) ∈ el(A,G);

SGK,k(eGH,h) =

{1 if (H, h) =G (K, k)

0 otherwise.

(v)

CB(A,G) =⊕

(V,ν)∈Gch(A,G)

C[AνG/V ] =⊕

(H,h)∈Gel(A,G)

CeGH,h.

(vi) For any z ∈ CB(A,G),

z =∑

(H,h)∈Gel(A,G)

SGH,h(z)eGH,h.

(vii) For any z ∈ CB(A,G) and (H, h) ∈ el(A,G), zeGH,h = SGH,h(z)eGH,h.

Proof : See [1].

�

Remark 5.1 was obtained by Barker in [1]. It is very important and used through-

out in the next chapters without referring sometimes.

Chapter 6

Some Maps

There are certain important maps defined in [16] for the Burnside rings. To

realize B(A,G) as a Mackey functor, in this chapter we extend these maps to

B(A,G) which contains B(G) as a unital subring, and we find the images of the

primitive idempotents of CB(A,G) under all these maps, except one, namely, the

orbit map.

If A is taken to be the trivial group, then our results recover the corresponding

results about these maps defined on B(G).

We are still assuming that A is a finite abelian group. However, for the places

in which the algebra maps SGH,h or the primitive idempotents eGH,h appear we have

to assume that A is a finite cyclic group regarded as a subgroup of C∗. Moreover,

wherever SGH,h or eGH,h appear, it must be understood that we extended these maps

by C−linear extension from B(A,G) to CB(A,G).

In fact, there are six maps that we want to consider. One of them , namely,

conjugation map, is very trivial. Two of them were studied, and the images of the

primitive idempotents under these two maps were found by Barker in [1]. Hence,

for these three maps we just state the results without proofs.

30

CHAPTER 6. SOME MAPS 31

6.1 The Inflation Map

Let N E G and S = AX be an A−fibred G/N−set. Define the inflated set

infGN (S) = S and let AG act on infGN (S) as;

(ag, s) 7→ ags = a(gN)s for all ag ∈ AG and s ∈ infGN (S).

Remark 6.1 Let S = AX and T = AY be A−fibred G/N−sets where N E G.

(i) infGN (S) is an A−fibred G−set.

(ii) S 'A(G/N) T if and only if infGN (S) 'AG infGN (T ).

(iii) infGN (S ] T ) = infGN (S) ] infGN (T ).

(iv) infGN (S ⊗ T ) = infGN (S)⊗ infGN (T ).

(v) S is a transitive A−fibred G/N−set if and only if infGN (S) is a transitive

A−fibred G−set.

Proof : (i) The action is well-defined: Suppose a1g1 = a2g2 ∈ AG and s1 = s2 ∈infGN (S) = S. We want to show that a1g1s1 = a2g2s2, equivalently (a1(g1N))s1 =

(a2(g2N))s2. Since a1(g1N) = a2(g2N) and s1 = s2 we have already (a1(g1N))s1 =

(a2(g2N))s2 because the action of A(G/N) on S is well defined.

The action properties are satisfied: Obvious.

infGN (S) is A−free: It is clear because the actions of A on S and infGN (S) are the

same.

(ii) Since infGN (S) = S and infGN (T ) = T, any bijective map from S to T is

a bijective map from infGN (S) to infGN (T ) and conversely. It is clear from the

definition of the AG−action on inflated sets that a bijective map from S to T

respects the A(G/N)−action if and only if it respects the AG−action.

(iii) and (iv) It is immediate because infGN (S) = S and infGN (T ) = T.

(v) It is clear by the action of AG on infGN (S).

�

Hence, by 6.1 we have a well-defined map, called the inflation map,

InfGN : B(A,G/N)→ B(A,G) given by InfGN ([S]) 7→ [infGN (S)] for any A−fibred

G/N−set S.

Remark 6.2 InfGN : B(A,G/N)→ B(A,G) is a ring monomorphism.


Proof : It is a well-defined map from 6.1(i) and (ii). It is a ring homomorphism

from (6.1)(iii) and (iv). Finally, injectivity follows from 6.1(ii).

�

Let N E G and N ≤ V ≤ G. For any ν ∈ Hom(V/N,A), we write ν for the

group homomorphism V → A given by ν(v) = ν(vN) for all v ∈ V.

Remark 6.3 Let N E G and (V/N, ν) ∈ ch(A,G/N). Then

InfGN ([Aν(G/N)

(V/N)]) = [AνG/V ].

Proof : InfGN ([Aν(G/N)(V/N)

]) = [infGN (Aν(G/N)(V/N)

)], and by 6.1(v)

infGN (Aν(G/N)(V/N)

) = Aν(G/N)(V/N)

is a transitive A−fibred G−set. Hence the AG−orbits

of its elements are all equal to Aν(G/N)(V/N)

.

Remember that Aν(G/N)(V/N)

= A(G/N)4(V/N,ν)

where 4(V/N,ν) = {ν((vN)−1)(vN) : vN ∈V/N} which is a subgroup of A(G/N). Put 4 = 4(V/N,ν).

We find the AG−stabilizer of 4 = 1(1N)4 ∈ A(G/N)4 :

Let ag ∈ AG. Then ag is in the stabilizer if and only if a(gN)4 = 4, which is

equivalent to g ∈ V and ν((gN)−1) = ν(g−1) = a. So, ag ∈ AG is in the stabilizer

if and only if ag ∈ {ν(g−1)g : g ∈ V } = 4(V,ν). Therefore;

infGN (AνA(G/N)V/N

) = orbAG(4) 'AG (AG)/4(V,ν) = AνG/V, as desired.

�

For the rest of this section, we consider the C−linear extension of the inflation

map; InfGN : CB(A,G/N)→ CB(A,G).

Lemma 6.4 Let N E G and (H, h) ∈ el(A,G). Then for any A−fibred G/N−set

S = AX we have;

SGH,h(InfGN ([S])) = S

G/N(NH)/N,hN([S]).

Proof : It suffices to prove this lemma for transitive A−fibred G/N−sets. Hence

take any (V/N, ν) ∈ ch(A,G/N). Remember that ν is the group homomorphism


from V to A given by ν(v) = ν(vN). Using 6.3;

SGH,h(InfGN ([Aν

(G/N)

(V/N)])) = SGH,h([AνG/V ]) =

∑gV⊆G,H≤gV

gν(h).

Note that; since N ≤ V, V/N = (NV )/N. So, H ≤ gV if and only if ((NH)/N) ≤gN((NV )/N) = gN(V ). Hence {gV ⊆ G : H ≤ gV } = {(gN)(V/N) ⊆ (G/N) :

((NH)/N) ≤ gN(V/N)}. Also gν(h) = gNν(hN). Therefore the last sum can be

written as; ∑gV⊆G,H≤gV

gν(h) =∑

(gN)(V/N)⊆(G/N),((NH)/N)≤gN (V/N)

gNν(hN)

= SG/N(NH)/N,hN([Aν

(G/N)

(V/N)]).

�

Theorem 6.5 Let N E G and (K/N, kN) ∈ el(A,G/N). Then

InfGN (eG/NK/N,kN) =

∑(H,h)∈Gel(A,G),((NH)/N,hN)=G/N (K/N,kN)

eGH,h.

Proof : For some complex numbers λH,h;

InfGN (eG/NK/N,kN) =

∑(H,h)∈Gel(A,G)

λH,heGH,h.

Then by 6.4;

λH,h = SGH,h(InfGN (e

G/NK/N,kN)) = S

G/N(NH)/N,hN(e

G/NK/N,kN).

Therefore

λH,h =

{1 , ((NH)/N, hN) =G/N (K/N, kN)

0 , otherwise.Thus, the result follows.

�


6.2 The Invariance Map

Let N E G and S = AX be an A−fibred G−set . Let invGN(S) = {s ∈ S :

Ns = s}. We let A(G/N) act on invGN(S) as: (a(gN), s) 7→ (a(gN))s = ags for

all a(gN) ∈ A(G/N) and s ∈ invGN(S).

Remark 6.6 Let N E G and S = AX be an A−fibred G−set. Then invGN(S) is

an A−fibred G/N−set.

Proof : invGN(S) is closed under the A(G/N)−action: Let a(gN) ∈ A(G/N)

and s ∈ invGN(S). We want to show that a(gN)s = ags ∈ inGN(S). For any n ∈ N,n(ags) = ag((g−1ng)s). By the normality of N, g−1ng ∈ N, and since s ∈ invGN(S)

is fixed by N we have n(ags) = ags implying that invGN(S) is closed under the

action of A(G/N).

invGN(S) is an A(G/N)−set: It is a straightforward checking of action properties.

invGN(S) is A−free: It is clear because the actions of A on invGN(S) and S are the

same.

�

Remark 6.7 Let S = AX and T = AY be A−fibred G−sets. Then

(i) If S 'AG T, then invGN(S) 'A(G/N) invGN(T ).

(ii) invGN(S ] T ) = invGN(S) ] invGN(T ).

(iii) If S is a transitive A−fibred G−set, then invGN(S) is a transitive A−fibred

G/N−set.

Proof : (i) Suppose S 'AG T. Then there is a bijective AG−map from S to T. It

is clear that the restriction of this map to invGN(S) yields a bijective A(G/N)−map

from invGN(S) to invGN(T ) which shows that invGN(S) 'A(G/N) invGN(T ).

(ii) and (iii) are obvious.

�

Hence by 6.6 and 6.7, we have a well-defined map, called the invariance map,

InvGN : B(A,G)→ B(A,G/N) given by InvGN([S]) = [invGN(S)] for any A−fibred

G−set S.


Remark 6.8 InvGN : B(A,G)→ B(A,G/N) is a Z−module homomorphism.

Proof : Follows from 6.6 and 6.7.

�

Let N E G and V ≤ G. For any ν ∈ Hom(V,A) such that N ≤ Kerν, we write ν

for the group homomorphism V/N → A given by ν(vN) = ν(v) for all vN ∈ V/N.Note that N ≤ Kerν implies that N ≤ V, and also that ν is well-defined.

Remark 6.9 Let N E G and (V, ν) ∈ ch(A,G). Then we have

InvGN([AνG/V ]) =

{[Aν

(G/N)(V/N)

], N ≤ Kerν

0, N � Kerν.

Proof : InvGN([AνG/V ]) = [invGN(AνG/V )]. Remember AνG/V = AG/4(V,ν)

where 4(V,ν) = {ν(v−1)v : v ∈ V } which is a subgroup of AG. Put 4(V,ν) = 4.Now, ag4 ∈ invGN(AνG/V ) if and only if ang4 = ag4 for all n ∈ N which is

equivalent to g−1ng ∈ 4 for all n ∈ N. Then by the definition of 4, g−1ng ∈ 4for all n ∈ N if and only if g−1ng ∈ V and ν((g−1ng)−1) = 1 for all n ∈ N that

is to say, n ∈ gV and gν(n−1) = 1 for all n ∈ N. Since N is normal, n ∈ gV andgν(n−1) = 1 for all n ∈ N if and only if N ≤ Kergν = g(Kerν), equivalently

N ≤ Kerν. Hence,

invGN(AνG/V ) =

{empty if N � Kerν

AνG/V if N ≤ Kerν.

So, if N � Kerν, then InvGN([AνG/V ]) = 0.

Suppose now N ≤ Kerν. Then invGN(AνG/V ) = AνG/V, and it is a transitive

A−fibred G/N−set by 6.7. Thus A(G/N)−orbits of its elements are all equal to

AνG/V.

We find the A(G/N)−stabilizer of 4 = 1.14 ∈ AG/4 :

a(gN) is in the stabilizer if and only if ag4 = 4, equivalently ag ∈ 4. Then

by the definition of 4, ag ∈ 4 if and only if g ∈ V and ν(g−1) = a. Using

N ≤ Kerν ≤ V, we see that a(gN) is in the stabilizer if and only if gN ∈ V/Nand ν((gN)−1) = a. Hence, stabA(G/N)(4) = {ν((gN)−1)(gN) : gN ∈ V/N} =

4(V/N,ν). Consequently for N ≤ Kerν;

invGN(AνG/V ) = AνG/V = orbA(G/N)(4) 'A(G/N)A(G/N)4(V/N,ν)

= Aν(G/N)(V/N)

. So we


proved the desired result.

�

Remark 6.10 Suppose N E G with O(N) = N and N ≤ V ≤ G. Then for any

ν ∈ Hom(V,A), N ≤ Kerν.

Proof : The restriction of ν to N is a group homomorphism from N to A. So

O(N) ≤ Kerν.

�

For the rest of this section we consider C−linear extension of the invariance map;

InvGN : CB(A,G)→ CB(A,G/N).

Lemma 6.11 Let N E G with O(N) = N and (H/N, hN) ∈ el(A,G/N). Then

for any A−fibred G−set S = AX we have;

SG/NH/N,hN(InvGN([S])) = SGH,h([S]).

Proof : It suffices to show this lemma for transitive A−fibred G−sets. Hence,

take any (V, ν) ∈ ch(A,G). From 6.10, N ≤ Kerν if and only if N ≤ V. Also

remember that ν denotes the group homomorphism V/N → A given by ν(vN) =

ν(v) if N ≤ Kerν.

Case (1): N ≤ V.

Using 6.9 we have

SG/NH/N,hN(InvGN([AνG/V ])) = S

G/NH/N,hN([Aν

(G/N)

(V/N)])

=∑

(gN)(V/N)⊆(G/N),(H/N)≤gN (V/N)

gN ν(hN)

=∑

gV⊆G,H≤gV

gν(h) = SGH,h([AνG/V ]).

Case (2): N � V.

By 6.9, InvGN([AνG/V ]) = 0 and so SG/NH/N,hN(InvGN([AνG/V ])) = 0.

On the other hand SGH,h([AνG/V ]) =∑

gV⊆G,H≤gVgν(h) = 0 because the indices


of the last sum range in the set {gV ⊆ G : H ≤ gV } which is empty. Otherwise;

if H ≤ gV for some g ∈ G then from N ≤ H we get N ≤ gV, and so by

the normality of N we have N ≤ V which is not the case. So for N � V,

SG/NH/N,hN(InvGN([AνG/V ])) and SGH,h([AνG/V ]) are both equal to 0.

�

Theorem 6.12 Let N E G, O(N) = N and (K, k) ∈ el(A,G). Then

InvGN(eGK,k) =

{0 if N � K

eG/NK/N,kN if N ≤ K.

Proof : For some complex numbers λH/N,hN ;

InvGN(eGK,k) =∑

(H/N,hN)∈G/Nel(A,G/N)

λH/N,hNeG/NH/N,hN .

Using 6.11;

λH/N,hN = SG/NH/N,hN(InvGN(eGK,k)) = SGH,h(e

GK,k).

Hence,

λH/N,hN = SGH,h(eGK,k) =

{1, (H, h) =G (K, k)

0, otherwise.Therefore;

InvGN(eGK,k) =∑

(H/N,hN)∈G/Nel(A,G/N),(H,h)=G(K,k)

eG/NH/N,hN .

The condition (H, h) =G (K, k) implies that N ≤ H = gK for some g ∈ G, and

by the normality of N we get N ≤ K. Hence, InvGN(eGK,k) = 0 if N � K.

Suppose now N ≤ K. If (H1, h1) =G (K, k) =G (H2, h2) where H1 ≥ N ≤ H2,

then (H1/N, h1N) =G/N (K/N, kN) =G/N (H2/N, h2N) (O(N) = N is used here!

Note that in this case N ≤ T implies that O(T/N) = O(T )/N). So the last sum

can not contain more than one summand, and clearly (K/N, kN) is a summand.

Hence, InvGN(eGK,k) = eG/NK/N,kN if N ≤ K.

�

Note that InvGN is not a multiplicative map in general (in contrast to the in-

variance map defined on B(G)). Because, for example, it may happen that there


are ν ∈ Hom(G,A) and ω ∈ Hom(G,A) such that N � Kerν, N � Kerω but

N ≤ Ker(ν.ω). In this case; InvGN([Aν.ωG/G]) 6= 0 but

[AνG/G][AωG/G] = [Aν.ωG/G], InvGN([AνG/G])InvGN([AωG/G]) = 0.

Remark 6.13 If O(N) = N, then InvGN is multiplicative and so a ring homo-

morphism.

Proof : Take any two elements x, y ∈ CB(A,G) where, say,

x =∑

(H,h)∈Gel(A,G)

λH,heGH,h, y =

∑(K,k)∈Gel(A,G)

µK,keGK,k.

Then using 6.12;

InvGN(x) =∑

(H,h)∈Gel(A,G),N≤H

λH,heG/NH/N,hN ,

InvGN(y) =∑

(K,k)∈Gel(A,G),N≤K

µK,keG/NK/N,kN .

Thus,

InvGN(x)InvGN(y) =∑

(T,t)∈Gel(A,G),N≤T

λT,tµT,teG/NT/N,tN .

On the other hand;

xy =∑

(T,t)∈Gel(A,G)

λT,tµT,teGT,t,

implying from 6.12 that

InvGN(xy) =∑

(T,t)∈Gel(A,G),N≤T

λT,tµT,teG/NT/N,tN .

Consequently, InvGN(xy) = InvGN(x)InvGN(y).

�

Lastly, if A is taken to be the trivial group then B(A,G) = B(G) and O(N) = N

for any subgroup N of G and so our invariance map extends the invariance map

defined for B(G) in [16].


6.3 The Orbit Map

First we must recall some facts from chapter 2. For any A−fibred G−set S =

AX and s ∈ S, νs is the uniquely determined element of Hom(stabG(As), A)

by the condition: gs = νs(g)s for all g ∈ stabG(As). Moreover, orbAG(s) 'AGAG/4(stabG(As),νs) = AνsG/stabG(As) where 4(stabG(As),νs) = {νs(g−1)g : g ∈stabG(As)} ≤ AG.

Let N E G and S be a G−set. Let S \ N be the set of N−orbits of S.

So, S \ N = {orbN(s) : s ∈ S}. S \ N becomes a G/N−set with the action:

(gN, orbN(s)) 7→ (gN)(orbN(s)) = orbN(gs).

To extend this definition to A−fibred G−sets, one can attempt to take all

N−orbits of an A−fibred G−set S = AX. Then it becomes an A(G/N)−set,

however it may not be A−free.

Let N E G and S = AX be an A−fibred G−set. Define

S \ \N = {orbN(s) : s ∈ S, stabG(As) ∩N ≤ Kerνs}.

We let A(G/N) act on S \ \N as: (a(gN), orbN(s)) 7→ (a(gN))(orbN(s)) =

orbN(ags) for all a(gN) ∈ A(G/N) and orbN(s) ∈ S \ \N.

Remark 6.14 For any A−fibred G−set S = AX, S \ \N is an A−fibred

G/N−set.

Proof : If orbN(s) ∈ S\\N and a(gN) ∈ A(G/N), then orbN(ags) ∈ S\\N : We

want to show that stabG(Aags)∩N ≤ Kerνags. We have stabG(As)∩N ≤ Kerνs.

From chapter 2 we know that stabG(Aags) = stabG(Ags) = gstabG(As) and

νags = νgs = gνs. Hence by the normality of N, stabG(As) ∩ N ≤ Kerνs implies

stabG(Ags) ∩N ≤ Kerνags.

If orbN(s1) = orbN(s2) ∈ S \ \N and a1(g1N) = a2(g2N) ∈ A(G/N), then

orbN(a1g1s1) = orbN(a2g2s2) :

orbN(s1) = orbN(s2) implies that ns1 = s2 for some n ∈ N. Also from a1(g1N) =

a2(g2N) it follows that a1 = a2 and g1n′ = g2 for some n′ ∈ N. Now, a2g2s2 =


a1g1n′ns1 = a1(g1n

′ng−11 )g1s = (g1n

′ng−11 )a1g1s1. Since g1n

′ng−11 ∈ N (N is

normal), orbN(a1g1s1) = orbN(a2g2s2).

The action properties are satisfied: It is obvious.

S \ \N is A−free: For any orbN(s) ∈ S \ \N we find its A−stabilizer as follows.

If a is in the A−stabilizer of orbN(s), then orbN(s) = orbN(as). Then ns = as

for some n ∈ N and so n(As) = A(as) = As implying that n ∈ stabG(As). But

we have already n ∈ N. So now n ∈ stabG(As) ∩ N ≤ Kerνs implying that

as = ns = νs(n)s = s. Because S is A−free, as = s implies a = 1. Thus, S \ \Nis A−free.

Being an A−free A(G/N)−set, S \ \N is an A−fibred G/N−set.

�

Remark 6.15 Let S = AX and T = AY be A−fibred G−sets. Then

(i) If S 'AG T then (S \ \N) 'A(G/N) (S \ \N).

(ii) (S ] T ) \ \N = (S \ \N) ] (T \ \N).

(iii) If S is a transitive A−fibred G−set, then S \ \N is a transitive A−fibred

G/N−set.

Proof : (i) Suppose S 'AG T. Then there is a bijective AG−map f : S → T.

Define f : S \ \N → T \ \N with the rule f(orbN(s)) = orbN(f(s)) for all

orbN(s) ∈ S \ \N.(1) Since f is bijective and preserves the AG−action, we have stabG(As) =

stabG(Af(s)) and νs = νf(s). Hence, orbN(s) ∈ S \ \N implies that orbN(f(s)) ∈T \ \N.(2) Suppose f(orbN(s)) = f(orbN(s′)) for some orbN(s) and orbN(s′) in S \ \N.Then orbN(f(s)) = orbN(f(s′)) and so nf(s) = f(s′) for some n ∈ N. Since f

respects the AG−action, nf(s) = f(s′) implies that f(ns) = f(s′). Then ns = s′

by the injectivity of f, and so orbN(s) = orbN(s′). Thus f is injective.

(3) f is surjective because f is surjective.

(4) It is clear that f preserves the A(G/N)−action because f preserves the

AG−action.

Hence we proved that (S \ \N) 'A(G/N) (T \ \N).

(ii) Obvious.


(iii) Suppose S is a transitive A−fibred G−set. Take any two elements orbN(s)

and orbN(s′) from S \ \N. Since S is transitive, there is an ag ∈ AG such that

ags = s′. But then (a(gN))orbN(s) = orbN(s′) implying that S\\N is a transitive

A−fibred G/N−set.

�

Hence, by 6.15 we have a well-defined map, called the orbit map,

OrbGN : B(A,G) → B(A,G/N) given by OrbGN([S]) = [S \ \N ] for all A−fibred

G−set S.

Remark 6.16 OrbGN : B(A,G)→ B(A,G/N) is a Z−module homomorphism.

Proof : It follows from 6.15.

�

Remark 6.17 For any (V, ν) ∈ ch(A,G);

(AνG/V ) \ \N =

{empty if V ∩N � Kerν

consist of all N−orbits if V ∩N ≤ Kerν.

Proof : Remember that AνG/V = AG/4(V,ν) where 4(V,ν) = {ν(v−1)v : v ∈V }. Put 4 = 4(V,ν). In chapter 2 we showed that stabG(Aag4) = gV and

νag4 = gν. Hence;

(AνG/V ) \ \N = {orbN(ag4) : ag ∈ AG, gV ∩N ≤ Ker(gν)}.

Since Ker(gν) = g(Kerν), using the normality of N we get

(AνG/V ) \ \N = {orbN(ag4) : ag ∈ AG, V ∩N ≤ Kerν}

which proves the desired result.

�

Let N E G and V ≤ G. For any ν ∈ Hom(V,A) such that V ∩ N ≤ Kerν, we

write ν for the group homomorphism (NV )/N → A given by ν(nvN) = ν(v) for

all nvN ∈ (NV )/N. Note that ν is well-defined.


Remark 6.18 Let N E G and (V, ν) ∈ ch(A,G). Then we have

OrbGN([AνG/V ]) =

{[Aν

(G/N)((NV )/N)

], V ∩N ≤ Kerν

0, V ∩N � Kerν.

Proof : If V ∩N � Kerν, from 6.17 OrbGN([AνG/V ]) = 0.

Suppose V ∩ N ≤ Kerν. OrbGN([AνG/V ]) = [(AνG/V ) \ \N ] is a transitive

A−fibred G/N−set from 6.15(iii). Also from 6.17, orbN(4) ∈ (AνG/V ) \ \Nwhere 4 = 1.14 ∈ AG/4 = AνG/V and 4 = {ν(v−1)v : v ∈ V }. So by the

transitivity of (AνG/V ) \ \N ;

(AνG/V ) \ \N = orbA(G/N)(orbN(4)) 'A(G/N)A(G/N)

stabA(G/N)(orbN (4)). Now, a(gN) ∈

stabA(G/N)(orbN(4)) if and only if orbN(ag4) = orbN(4) which is to say that

ag4 = n4 for some n ∈ N. But then the definition of4 yields; ag4 = n4 if and

only if n−1g ∈ V and ν((n−1g)−1) = a, or equivalently a(gN) ∈ {ν((gN)−1)(gN) :

gN ∈ (NV )/N} = 4((NV )/N,ν). Thus,

(AνG/V ) \ \N 'A(G/N)A(G/N)

4((NV )/N,ν)= Aν

(G/N)(NV )/N

.

�

For the orbit map, we could not find the images of the primitive idempotents

of CB(A,G).

There are three more maps, for two of which the images of primitive idempo-

tents of CB(A,G) were found in [1]. The remaining one is the conjugation map

that is very trivial. For this reason in the following three sections we give the

definitions and results without proving them.

6.4 The Conjugation Map

Let F ≤ G and g ∈ G. For any A−fibred F−set S = AX, we define g−conjugategS of S as the A−fibred gF−set gS = S with the AgF -action: (agf)s = (af)s

for all agf ∈ AgF and s ∈ gS. Then we have a well-defined map, called the

conjugation map,


CongF : B(A,F )→ B(A, gF ) given by CongF ([S]) = [gS] for any A−fibred F−set

S.

Remark 6.19 (i) CongF : B(A,F )→ B(A, gF ) is a Z−algebra isomorphism.

(ii) CongF ([AνF/V ]) = [AgνgF/gV ].

(iii) CongF (eFK,k) = egFgK,gk.

Proof : Obvious.

�

6.5 The Restriction Map

Let F ≤ G, and S = AX be an A−fibred G−set. By restricting the AG−action

to AF, we get an A−fibred F−set resGF (S) = S. Then we have a well-defined

map, called the restriction map,

ResGF : B(A,G) → B(A,F ) given by ResGF ([S]) = [resGF (S)] for all A−fibred

G−set S = AX.

Remark 6.20 (i) ResGF : B(A,G)→ B(A,F ) is a Z−algebra homomorphism.

(ii)

ResGF ([AνG/V ]) =∑

FgV⊆G

[AgνF/F ∩ gV ].

(iii)

ResGF (eGK,k) =∑

(H,h)∈F el(A,F ),(H,h)=G(K,k)

eFH,h.

Proof : See [1].

�


6.6 The Induction Map

Let H ≤ G and S = AX be an A−fibred H−set. Then the cartesian prod-

uct AG × S becomes an A−fibred H−set with the action: (ah, (bg, s)) 7→(a−1bh−1g, ahs) for all ah ∈ AH, bg ∈ AG and s ∈ S.Let AG×AH S denote the set of AH−orbits of the AH−set AG× S. That is,

AG×AH S = {orbAH(ag, s) : (ag, s) ∈ AG× S}.

Then AG×AH S becomes an A−fibred G−set with the AG−action:

(ag, orbAH(bh, s)) 7→ orbAH(abgh, s) for all orbAH(bh, s) ∈ AG ×AH S and ag ∈AG.

So we have a well-defined map, called the induction map,

IndGH : B(A,H)→ B(A,G) given by IndGH([S]) = [AG×AH S] for any A−fibred

H−set S = AX.

Remark 6.21 (i) IndGH : B(A,H)→ B(A,G) is a Z−module homomorphism.

(ii) IndGH([AνH/V ]) = [AνG/V ].

(iii)

IndGH(eHK,k) =|NG(K, k)||NH(K, k)|

eGK,k.

Proof : Remember that el(A,G) is a G−set by conjugation. The notations in

(iii) are NH(K, k) = stabH((K, k)) and NG(K, k) = stabG((K, k)). For the proof

see [1].

�

Chapter 7

Prime Ideals Of B(A,G)

In this chapter we will find the prime ideals of the monomial Burnside rings and

try to get some consequences about the primitive idempotents of the monomial

Burnside rings tensored over Z with an integral domain of characteristic 0. Some

of the results we are going to obtain are already obtained by Dress [9] and Barker

[1].

In [1], Barker found the primitive idempotents of CB(A,G) and gave a formula

expressing the primitive idempotents of CB(A,G) in terms of the transitive basis

of CB(A,G).

In [9], Dress found the prime spectrum of the monomial Burnside rings and gave

some consequences including his celebrated characterization of solvable groups.

As stated in the introduction Dress introduced the monomial Burnside rings in [9]

but his monomial Burnside ring is more general than the one we are considering

here. Because of the full generality of the assumptions in [9], we follow a different

approach some of whose parts inspired from [9].

Let ζ be a primitive nth root of unity and A be a cyclic group of order n.

So we can assume that A ≤ D∗ where D = Z[ζ]. We first find the prime ideals

of DB(A,G). Some of the results that we will obtain for DB(A,G) hold for

RB(A,G) where R is a ring which is more general than D. After finding the

prime ideals and the spectrum of DB(A,G), and some consequences we state (if

possible) our results for RB(A,G) where R is a ring satisfying weaker assumptions

45

CHAPTER 7. PRIME IDEALS OF B(A,G) 46

than D satisfies. To do so we need to remember some theory about integral ring

extensions, prime ideals of product rings and Dedekind domains. Our rings will

be commutative with identity elements.

A ring extension of a ring R is a ring S containing R as a unital subring. We

write R ≤ S to denote that S is a ring extension of R.

Let R ≤ S be a ring extension, an element s ∈ S is said to be integral over R if

f(s) = 0 for some monic polynomial f(x) ∈ R[x].

A ring extension R ≤ S is said to be integral if every element of S is integral over

R.

The following six theorems are all well-known (See [11] and [13]).

Theorem 7.1 Let R ≤ S be a ring extension and s ∈ S. The following condi-

tions are equivalent;

(i) s is integral over R.

(ii) R[s] is a finitely generated R−module.

(iii) There exists a faithful R[s]−module which is finitely generated R−module.

Theorem 7.2 Let R ≤ S be a ring extension. If S is finitely generated

R−module, then R ≤ S is an integral extension.

Theorem 7.3 Let R ≤ S be a ring extension. For any prime ideal P of S, R∩Pis a prime ideal of R.

For a ring extension R ≤ S, not all prime ideals of R is of the form R ∩ P where

P is a prime ideal of S. However if R ≤ S is an integral extension then prime

ideals of R are of the form R ∩ P where P is a prime ideal of S. Indeed,

Theorem 7.4 Let R ≤ S be an integral ring extension. Then we have;

(i)(Lying Over) Let P be a prime ideal of R. Then for any ideal I of S such that

R ∩ I ⊆ P there exists a prime ideal Q of S such that I ⊆ Q and R ∩ Q = P .

In particular, for any prime ideal P of R there exists a prime ideal Q of S with

R ∩Q = P .


(ii)(Going Up) Given prime ideals P $ P0 of R and Q of S with R ∩ Q = P ,

there exists a prime ideal Q0 of S satisfying Q $ Q0 and R ∩Q0 = P0.

(iii)(Going Down) Given prime ideals P % P0 of R and Q of S with R∩Q = P ,

there exists a prime ideal Q0 of S satisfying Q % Q0 and R ∩Q0 = P0.

(iv)(Incomparability) Two different prime ideals of S having the same intersec-

tion with R cannot be comparable.

(v)(Maximality) Let P be a prime ideal of R and Q of S with R ∩Q = P . Then

Q is a maximal ideal of S if and only if P is a maximal ideal of R.

(vi)(Minimality) Let P be a prime ideal of R and Q of S with R ∩Q = P . Then

Q is a minimal prime ideal of S if and only if P is a minimal prime ideal of R.

Theorem 7.5 Let Ri, i ∈ I, be a family of rings and πi be the projection of the

product ring∏

i∈I Ri onto the ith term Ri. Then any prime ideal of∏

i∈I Ri is

of the form π−1i (P ) for some i ∈ I and prime ideal P of Ri. Moreover, for any

i ∈ I and prime ideal P of Ri π−1i (P ) is a prime ideal of

∏i∈I Ri.

An integral domain R is called Dedekind domain if the following three conditions

hold:

(1) R is Noetherian,

(2) R is integrally closed,

(3) Every nonzero prime ideal of R is maximal.

Theorem 7.6 Let R be a Dedekind domain. Then

(i) If I is an ideal of R, then R/I is a principal ideal ring,

(ii) For any ideal I of R and nonzero a ∈ I, there exists b ∈ I such that I = (a, b),

(iii) For any prime ideal P of R the localized ring RP is a principal ideal domain.

Let K = Q[ζ] and we are assuming that A ≤ D∗ where D = Z[ζ]. Note that D

is a Dedekind domain and K is its field of fractions. For any (H, h) ∈G el(A,G)

we have a ring epimorphism

SGH,h : DB(A,G) −→ D given by [AνG/V ] 7→∑

gV⊆G,H≤gV

gν(h),


and the product map ψ =∏

(H,h)∈Gel(A,G) SGH,h is injective. For any nonzero

prime ideal P of D let πP be the canonical epimorphism from D to D/P . We

write I(H, h, 0) and I(H, h, P ) for the kernels of the maps SGH,h and πP ◦ SGH,hrespectively. Hence,

I(H, h, 0) = {x ∈ DB(A,G) : SGH,h(x) = 0},

I(H, h, P ) = {x ∈ DB(A,G) : SGH,h(x) ∈ P}.

Remark 7.7 Let P be a nonzero prime ideal of D and (H, h) ∈G el(A,G). Then

(i) DB(A,G)/I(H, h, 0) ' D,

(ii) DB(A,G)/I(H, h, P ) ' D/P,

(iii) I(H, h, 0) and I(H, h, P ) are prime ideals of DB(A,G).

Proof : Because D and D/P are domains and the maps SGH,h, and πP ◦ SGH,hare epimorphisms, the result follows.

�

Theorem 7.8 (i) Any prime ideal of DB(A,G) is of the form I(H, h, 0) or

I(H, h, P ) for some nonzero prime ideal P of D and (H, h) ∈G el(A,G).

(ii) If P is a nonzero prime ideal of D, then I(H, h, P ) is a maximal ideal of

DB(A,G) for any (H, h) ∈G el(A,G).

(iii) For any (H, h) ∈G el(A,G), I(H, h, 0) is a minimal prime ideal of DB(A,G).

(iv) For any nonzero prime ideal P of D and (H, h) ∈G el(A,G) we have

I(H, h, P ) ∩D = P , I(H, h, 0) ∩D = 0 and I(H, h, 0) $ I(H, h, P ).

(v) For any (H, h), (K, k) ∈G el(A,G) and prime ideals (possibly 0) P,Q of D

I(H, h, P ) = I(K, k,Q) implies that P = Q.

(vi) Let (H, h), (K, k) ∈G el(A,G). Then

I(H, h, 0) = I(K, k, 0) if and only if (H, h) =G (K, k).

Proof : (i) We know that the map ψ is injective. Hence we can write

DB(A,G) ≤∏

(H,h)∈Gel(A,G)D. Since both rings are finitely generated as


D−modules, the ring extension DB(A,G) ≤∏

(H,h)∈Gel(A,G)D is an integral ex-

tension.

Consequently by 7.4; any prime ideal of DB(A,G) is of the form ψ−1(I) where I

is a prime ideal of the product ring∏

(H,h)∈Gel(A,G)D. Then by 7.5; I = π−1(H,h)(P )

for some (H, h) ∈G el(A,G) and prime ideal P of D. Thus any prime ideal of

DB(A,G) is of the form

ψ−1(π−1(H,h)(P )) = (SGH,h)

−1(P ).

Note that

if P = 0, then (SGH,h)−1(P ) = Ker(SGH,h) = I(H, h, 0), and

if P 6= 0, then (SGH,h)−1(P ) = Ker(πP ◦ SGH,h) = I(H, h, P ).

(ii) and (iii) Since D is a Dedekind domain, its nonzero prime ideals are maxi-

mal. Hence for any nonzero prime ideal P of D π−1(H,h)(P ) is a maximal ideal, and

π−1(H,h)(0) is a minimal prime ideal of

∏(H,h)∈Gel(A,G)D. Now from 7.4 ((Maximal-

ity) and (Minimality)), the results follow.

(iv) Let P be a nonzero prime ideal of D. Since I(H, h, P ) is the kernel of the

D−linear ring epimorphism

πP ◦ SGH,h : DB(A,G)→ D → D/P

we have P ⊆ I(H, h, P ) ∩ D. Note that I(H, h, P ) ∩ D is a prime ideal of D

because D ≤ DB(A,G). As D is a Dedekind domain, P is a maximal ideal of

D. Also it is clear that I(H, h, P ) ∩D 6= D. Thus, maximality of P implies that

I(H, h, P ) ∩D = P .

The ring extension D ≤ DB(A,G) is an integral extension and I(H, h, 0) is a

minimal prime ideal of DB(A,G). Then by 7.4(Minimality) I(H, h, 0) ∩D must

be a minimal prime ideal of D. Being Dedekind domain, 0 is the only minimal

prime ideal of D.

It is clear that I(H, h, 0) ⊆ I(H, h, P ). Because SGH,h is D−linear, SGH,h(P ) = P .

As a result, P ⊆ I(H, h, P ) but P * I(H, h, 0).

(v) By part (iv).

(vi) Suppose I(H, h, 0) = I(K, k, 0). Any primitive idempotents of CB(A,G) lies

inKB(A,G). So for eGH,h we can find a nonzero d ∈ D(=product of denominators)

such that deGH,h ∈ DB(A,G). Then SGH,h(deGH,h − d) = 0. Hence deGH,h − d ∈


I(H, h, 0) = I(K, k, 0). We must have SGK,k(deGH,h−d) = 0. However SGK,k(de

GH,h−

d) = 0 if and only if (H, h) =G (K, k) (otherwise it is equal to −d; but in D, −dcannot be equal to 0 since D is of characteristic 0).

�

Lemma 7.9 Let P be a nonzero prime ideal of D and (H, h), (K, k) ∈ el(A,G).

Then

I(H, h, 0) ⊆ I(K, k, P ) if and only if I(H, h, P ) = I(K, k, P ).

Proof : (⇒) Suppose I(H, h, 0) ⊆ I(K, k, P ). Take any x ∈ I(H, h, P ). Then

SGH,h(x) = d ∈ D = Z[ζ] and d ∈ P. Because SGH,h is a D−linear ring epimorphism

from DB(A,G) to D, we have SGH,h(x − d) = 0 and so x − d ∈ I(H, h, 0) ⊆I(K, k, P ). Then SGK,k(x−d) = SGK,k(x)−d ∈ P. As d ∈ P, we must have SGK,k(x) ∈P. Consequently, x ∈ I(K, k, P ). So we proved that I(H, h, P ) ⊆ I(K, k, P ). But

then maximality of I(H, h, P ) implies that I(H, h, P ) = I(K, k, P ).

(⇐) It is clear from 7.8(iv).

�

The next lemma states that the maps πP ◦ SGH,h and πP ◦ SGK,k are equal if and

only if their kernels are equal.

Lemma 7.10 For a nonzero prime ideal P of D let πP be the canonical epimor-

phism from D onto D/P . Suppose (H, h), (K, k) ∈ el(A,G) and I(H, h, P ) =

I(K, k, P ). Then πP ◦ SGH,h = πP ◦ SGK,k.

Proof : Since I(H, h, P ) = I(K, k, P ), for any x ∈ DB(A,G), it follows that

SGH,h(x) ∈ P if and only if SGK,k(x) ∈ P.Take any x ∈ DB(A,G) we know that both or none of SGH,h(x) and SGK,k(x) belong

to P . If they are in P , then πP ◦ SGH,h(x) and πP ◦ SGK,k(x) are both equal to 0.

Assume that they are not in P . Then write SGH,h(x) = y1 and SGK,k(x) = y2 for

some y1, y2 ∈ D. Because D/P is a field there exists a z ∈ D with z /∈ P such

that

(z + P )(y1 + P ) = (y2 + P ).


Now zy1 − y2 ∈ P and note that zx − y2 ∈ DB(A,G) with SGH,h(zx − y2) =

zy1 − y2 ∈ P . Thus we must also have SGK,k(zx − y2) = zy2 − y2 ∈ P . But then

from zy1 − y2 ∈ P and zy2 − y2 ∈ P we get

(zy1 − y2)− (zy2 − y2) = z(y1 − y2) ∈ P.

Since P is a prime ideal and z /∈ P it follows that y1 − y2 ∈ P . Therefore;

for any x ∈ DB(A,G) SGH,h(x)−SGK,k(x) ∈ P . That is to say πP ◦SGH,h = πP ◦SGK,k.

�

We will study the prime spectrum of DB(A,G), written as Spec(DB(A,G)).

So we begin by some facts about spectrum of commutative rings (see [7]). Let

R be a commutative ring. Spec(R) is defined to be the set of prime ideals of R.

For any ideal I of R we put V (I) = {P ∈ Spec(R) : I ⊆ P}. We can define a

topology on Spec(R) by calling sets of the form V (I) closed. The closure of a

point P ∈ Spec(R) is {P} := P = V (P ). Suppose R is a Noetherian ring. Then

two prime ideals P,Q of R are in the same connected component of Spec(R) if

and only if there is a sequence P1, ...., Pn of minimal prime ideals of R such that

P ∈ P1, Q ∈ Pn, Pi ∩ Pi+1 6= ∅ for i = 1, 2, ..., n− 1.

Being in the same connected component of Spec(R) forms an equivalence rela-

tion on Spec(R). We write P ∼ Q if the prime ideals P,Q of R are in the same

connected component of Spec(R).

For a Noetherian ring R, we have a bijective correspondence between the con-

nected components of Spec(R) and the primitive idempotents ofR. The connected

component of Spec(R) corresponding to a primitive idempotent e of R consists

of all prime ideals of R containing 1− e.

Recall that for H ≤ G, O(H) = ∩ν∈Hom(H,A)Kerν. For any automorphism

α of H and ν ∈ Hom(H,A), ν ◦ α ∈ Hom(H,A). So O(H) is a characteristic

subgroup of H. For any g ∈ G and ν ∈ Hom(H,A), gν ∈ Hom(gH,A) implying

that O(gH) = gO(H). We write H for H/O(H), h for hO(H) and N(H) for

NG(H)/H.


NG(H) acts on H by conjugation;

(gH, hO(H)) := gh 7→ ghg−1O(h), for gH ∈ NG(H) and h ∈ H.The above action of NG(H) on H respects the multiplication of H. That is;g ¯(h1h2) = (gh1)(

gh2).

Note that H ≤ NG(H) fixes H pointwise and so H is also N(H)−set with respect

to conjugation action.

Lemma 7.11 Let P be a nonzero prime ideal of D such that D/P has charac-

teristic p. For (H, h1), (H, h2) ∈ el(A,G) with h−11 h2 ∈ H is of p−power order,

we have I(H, h1, P ) = I(H, h2, P ).

Proof : Let ν ∈ Hom(H,A). Put ν(h1) = d1 ∈ A ≤ D∗ and ν(h2) = d2 ∈A ≤ D∗. Since h−1

1 h2 has p−power order in H, we must have dpm

1 = dpm

2 for

some natural number m. For any d ∈ D, let d = d + P = πP (d) where πP is the

canonical ring epimorphism from D to D/P. Because D/P has characteristic p,

0 = d1pm

− d2pm

= (d1 − d2)pm.

So, 0 = d1 − d2 since a field has no nonzero nilpotent elements. Thus we have

ν(h1) + P = ν(h2) + P for any ν ∈ Hom(H,A). Therefore

πP ◦ SGH,h1([AνG/V ]) = πP ◦ SGH,h2

([AνG/V ]) for any [AνG/V ] ∈ B(A,G). That

is; I(H, h1, P ) = I(H, h2, P ).

�

Let G be a group and p be a prime number. For any g ∈ G there are uniquely

determined elements gp and gp′ of G satisfying; g = gpgp′ = gp′gp, the order of gp is

a p−power, and the order of gp′ is not divisible by p. Indeed, let pkm be the order

of g where (p,m) = 1. Then there are integers u and v such that pku +mv = 1,

and so g = (gmv)(gpku) = (gp

ku)(gmv) where gp = gmv and gp′ = gpku.

By the uniqueness of such representations, for any H ≤ G and h ∈ H we have

hpO(H) = (hO(H))p and hp′O(H) = (hO(H))p′ . 7.11 implies that I(H, h, P ) =

I(H, hp′ , P ) for any nonzero prime ideal of D such that D/P has characteristic

p.

Lemma 7.12 Let P be a prime ideal (possibly 0) of D and (H, h) ∈ el(A,G).


Then the prime ideals I(H, h, P ) and I(H, 1, 0) are in the same connected com-

ponent of Spec(DB(A,G)).

Proof : Since H is an abelian group, it is the direct product of its Sylow sub-

groups;

H = Sp1(H)× ...× Spr(H).

Then h can be written uniquely as h = t1...tr where ti ∈ Spi(H) for i = 1, ..., r.

Chose prime ideals P1, ..., Pr of D such that D/Pi has characteristic pi. It is pos-

sible to choose such prime ideals because D = Z[ζ].

For i = 1, 2, ..., r − 1; (ti...tr)−1(ti+1...tr) ∈ Spi

(H). So by 7.11 we have

I(H, h, P1) = I(H, t2t3...tr, P1), I(H, t2t3...tr, P2) = I(H, t3t4...tr, P2),

I(H, t3...tr, P3) = I(H, t4...tr, P3), .......,

I(H, tr−1tr, Pr−1) = I(H, tr, Pr−1), I(H, tr, Pr) = I(H, 1, Pr).

Also note that for any prime ideals A,B (possibly 0) of D and (K, k) ∈ el(A,G)

we have I(K, k,A) ⊇ I(K, k, 0) ⊆ I(K, k,B).

Thus, I(K, k,A) and I(K, k,B) are in the same connected component of

Spec(DB(A,G)). Recall that being in the same connected component is an equiv-

alence relation for which we use the notation ∼ . Now what we have is

I(H, h, P ) ∼ I(H, h, P1) = I(H, t2t3...tr, P1) ∼ I(H, t2t3...tr, P2) =

I(H, t3t4...tr, P2) ∼ I(H, t3...tr, P3) = I(H, t4...tr, P3) ∼

I(H, t4...tr, P4) = I(H, t5...tr, P4) ∼ I(H, t5...tr, P5) = .........

∼ I(H, tr−1tr, Pr−1) = I(H, tr, Pr−1) ∼ I(H, tr, Pr) = I(H, 1, Pr) ∼ I(H, 1, 0).

Hence, I(H, h, P ) ∼ I(H, 1, 0).

�


teristic p. Let H E K be such that K/H is a p−group. Then we have

I(H, 1, P ) = I(K, 1, P ).


Proof : Take any [AνG/V ] ∈ B(A,G) and compute;

SGH,1([AνG/V ]) =∑

gV⊆G,H≤gV

1; SGK,1([AνG/V ]) =∑

gV⊆G,K≤gV

1.

For any G−set S and H ≤ G, let SH denote the H−fixed points of S. We have

SGH,1([AνG/V ])− SGK,1([AνG/V ]) = |G/V H −G/V K |.

The set G/V H −G/V K is a K/H−set whose K/H−orbits are nontrivial. Since

K/H is a p−group, |G/V H −G/V K | is divisible by p. Thus,

πP ◦ SGH,1([AνG/V ])− πP ◦ SGK,1([AνG/V ]) = 0

where πP is the canonical ring epimorphism from D to D/P. Therefore πP ◦SGH,1 =

πP ◦ SGK,1 implying that I(H, 1, P ) = I(K, 1, P ).

�

Let H E K with K/H is a p−group. For any prime ideals A and B (possibly

0) of D we have I(H, 1,A) ∼ I(K, 1,B). Because; choosing a prime ideal P of D

such that D/P has characteristic p we have by 7.13

I(H, 1,A) ∼ I(H, 1, P ) = I(K, 1, P ) ∼ I(K, 1,B).

For a group G, let S(G) denote the unique minimal normal subgroup of G such

that G/S(G) is solvable. Indeed; S(G) is the intersection of all normal subgroups

of G with solvable quotient groups. Thus, S(G) is a characteristic subgroup of

G.

For g ∈ G and H ≤ G, we have S(gH) = gS(H) and S(S(H)) = S(H).

A group G is called perfect if S(G) = G, equivalently if D(G) = G where D(G)

is the commutator subgroup of G.


Then the prime ideals I(H, h, P ) and I(S(H), 1, 0) of DB(A,G) are in the same

connected component of Spec(DB(A,G)).

Proof : Because H/S(H) is solvable, it has a subnormal series whose factor

groups have prime orders;

H = H1 D H2 D H3 D ... D Hn−1 D Hn = S(H),


where for each i |Hi/Hi+1| =a prime number pi. Since D = Z[ζ], we can choose

prime ideals P1, ..., Pn of D such that D/Pi has characteristic pi.

(A): By 7.13;

I(H, 1, P1) = I(H2, 1, P1), I(H2, 1, P2) = I(H3, 1, P2),

I(H3, 1, P3) = I(H4, 1, P3), I(H4, 1, P4) = I(H5, 1, P4), ......,

I(Hn−2, 1, Pn−2) = I(Hn−1, 1, Pn−2),

I(Hn−1, 1, Pn−1) = I(Hn, 1, Pn−1) = I(S(H), 1, Pn−1).

(B): By 7.12;

I(H, h, P ) ∼ I(H, 1, 0) ∼ I(H, 1, P1), I(S(H), 1, Pn−1) ∼ I(S(H), 1, 0).

(C): Also we know already that I(K, k,A) ∼ I(K, k,B) for any prime ideals A,B

of D and (K, k) ∈ el(A,G).

Now since ∼ is an equivalence relation on Spec(DB(A,G)), it follows from (A),

(B) and (C) that I(H, h, P ) ∼ I(S(H), 1, 0).

�

Theorem 7.15 Let P and Q are prime ideals (possibly 0) of D and (H, h),

(K, k) ∈ el(A,G) with S(H) =G S(K). Then

the prime ideals I(H, h, P ) and I(K, k,Q) of DB(A,G) are in the same connected

component of Spec(DB(A,G)).

Proof : By 7.14;

I(H, h, P ) ∼ I(S(H), 1, 0), I(K, k,Q) ∼ I(S(K), 1, 0).

But S(H) =G S(K) implies that I(S(H), 1, 0) = I(S(K), 1, 0).

Since ∼ is an equivalence relation on Spec(DB(A,G)) we have

I(H, h, P ) ∼ I(K, k,Q).

�

For a group G and prime number p, let Op(G) be the unique minimal normal

subgroup of G such that G/Op(G) is a p−group. Actually, Op(G) is the inter-

section of all normal subgroups of G having p−power indexes. So Op(G) is a


characteristic subgroup of G.

Since p−groups are solvable, for a group G we have S(G) E Op(G).

Being a subgroup of the solvable group G/S(G), Op(G)/S(G) is solvable, and

so S(Op(G)) E S(G). As S(G) is a normal subgroup of G and S(Op(G)) is a

characteristic subgroup of S(G), we must have S(Op(G)) E G. Now the group

G/S(Op(G)) has a solvable normal subgroup Op(G)/S(Op(G)) whose quotient

group (isomorphic to G/Op(G)) is also solvable. Consequently, G/S(Op(G)) is

solvable and S(G) ≤ S(Op(G)). Therefore, S(Op(G)) = S(G) for any group G.

Let H E K be such that K/H is a p−group. Because Op(H) is a characteristic

subgroup of H and H is a normal subgroup of K, we get Op(H) E K. Then

K/Op(H) is a p−group and so Op(K) E Op(H). On the other hand; Op(K) is

a normal subgroup of H of p−power index. Thus Op(H) E Op(K). As a result,

Op(H) = Op(K) for any normal subgroup H of K of p−power index.


teristic p, and (H, h), (K, k) ∈ el(A,G). Suppose |NG(H) : H| and |NG(K) : K|are not divisible by p. Then

I(H, h, P ) ∩B(G) = I(K, k, P ) ∩B(G) implies that H =G K.

Proof : For any T ≤ G, let τ denote the trivial elements of the groups

Hom(T,A).

For [AτG/H], [AτG/K] ∈ B(G) we compute;

SGH,h([AτG/H]) =∑

gH⊆G,H≤gH

1 = |NG(H) : H|,

SGK,k([AτG/K]) =∑

gK⊆G,K≤gK

1 = |NG(K) : K|.

Since |NG(H) : H| and |NG(K) : K| are not divisible by p;

[AτG/H] /∈ I(H, h, P ), [AτG/K] /∈ I(K, k, P ).

Then from I(H, h, P ) ∩B(G) = I(K, k, P ) ∩B(G) we have;

[AτG/H] /∈ I(K, k, P ), [AτG/K] /∈ I(H, h, P ).


Hence in particular;

SGK,k([AτG/H]) =∑

gH⊆G,H≤gK

1 6= 0, SGH,h([AτG/K]) =∑

gK⊆G,K≤gH

1 6= 0.

Therefore, we must have H ≤G K ≤G H.

�


Then; I(H, h, P ) ∩B(G) = I(H, 1, P ) ∩B(G).

Proof : Let τ be the trivial elements of the groups Hom(T,A) for all T ≤ G.

Then for any [AτG/T ] ∈ B(G), SGH,h([AτG/T ]) and SGH,1([AτG/T ]) are both equal

to G/TH where G/TH is the set of H−fixed points of the G−set G/T. So the

result follows.

�

Theorem 7.18 Let P be a nonzero prime ideal of D such that D/P has char-

acteristic p, and (H, h), (K, k) ∈ el(A,G).

If I(H, h, P ) ∩B(G) = I(K, k, P ) ∩B(G), then Op(H) =G Op(K).

Proof : For any subgroup T of G we define a subnormal series;

T = T0 E T1 E T2... E Tn E ...

where Tn+1/Tn is a Sylow p−subgroup of N(Tn) = NG(Tn)/Tn for each n. Since

we are considering only finite groups this series must stop in a finite number,

say n. Then Tn+k = Tn for each natural number k, and so |NG(Tn) : Tn| is not

divisible by p.

For i = 1, 2, ..., n − 1; Ti E Ti+1 and Ti+1/Ti is a p−group. Hence by 7.13;

I(T, 1, P ) = I(Tn, 1, P ).

So finding the above series of H and K, we have

H = H0 E H1 E ... E Hr; K = K0 E K1 E ... E Ks.


Then I(H, 1, P ) = I(Hr, 1, P ), I(K, 1, P ) = I(Ks, 1, P ) implying by 7.17 that

I(Hr, 1, P ) ∩B(G) = I(Ks, 1, P ) ∩B(G).

Since |NG(Hr) : Hr| and |NG(Ks) : Ks| are not divisible by p, 7.16 gives that

Hr =G Ks.

Recall also that, for groups M E N with N/M is a p− group we have Op(M) =

Op(N).

Thus Op(H) = Op(H1) = ... = Op(Hr) =G Op(Ks) = ... = Op(K1) = Op(K).

�

Theorem 7.19 Let P and Q be prime ideals (possibly 0) of D and

(H, h), (K, k) ∈ el(A,G). If the prime ideals I(H, h, P ) and I(K, k,Q) of

DB(A,G) are in the same connected component of Spec(DB(A,G)), then

S(H) =G S(K).

Proof : By 7.14;

I(H, h, P ) ∼ I(S(H), 1, 0) and I(K, k,Q) ∼ I(S(K), 1, 0).

Since ∼ is an equivalence relation, I(S(H), 1, 0) ∼ I(S(K), 1, 0).

So (DB(A,G) is Noetherian as it is finitely generated Z−module) there is a se-

quence of minimal prime ideals

I(T1, t1, 0), ..., I(Tn, tn, 0) of DB(A,G) such that

(A) I(S(H), 1, 0) ⊇ I(T1, t1, 0),

(B) I(S(K), 1, 0) ⊇ I(Tn, tn, 0),

(C) Closures of the points I(Ti, ti, 0) and I(Ti+1, ti+1, 0) of Spec(DB(A,G)) in-

tersect nontrivially for each i = 1, 2, ..., n− 1.

From (A), (B) and 7.8(vi) it follows that (S(H), 1) =G (T1, t1) and

(S(K), 1) =G (Tn, tn). In particular, S(H) =G T1 and S(K) =G Tn. Hence

S(H) =G S(T1) and S(K) =G S(Tn).

From (C); for each i there is a nonzero prime ideal Ri of D and (Li, li) ∈ el(A,G)

such that

I(Ti, ti, 0) ⊆ I(Li, li, Ri) ⊇ I(Ti+1, ti+1, 0).

By 7.9, for each i = 1, 2, ...n− 1 we have

I(Ti, ti, Ri) = I(Ti+1, ti+1, Ri), and it implies from 7.18 that Opi(Ti) =G Opi(Ti+1)

where pi is the characteristic of D/Ri. But for any group E and prime number p,


S(Op(E)) = S(E).

Hence, S(H) =G S(T1) =G S(T2) =G ... =G S(Tn−1) =G S(Tn) =G S(K).

�

Consider the set F = {S(H) : H ≤ G}. Let =G denote the conjugacy relation

on the set of subgroups of G. We know from 7.15 and 7.19 that prime ideals

I(H, h, P ) and I(K, k,Q) of DB(A,G) are in the same connected component of

Spec(DB(A,G)) if and only if S(H) =G S(G). Hence, the number of connected

components of Spec(DB(A,G)) is equal to the |F/=G| = the number of noncon-

jugate perfect subgroups of G. F has only one nonconjugate element if and only

if G is solvable. We state some simple consequences of what we proved up to

this point in the following corollary which contains also the characterization of

solvable groups given first by Dress in [9].

Corollary 7.20 (i) The connected components of Spec(DB(A,G)) are in bijec-

tive correspondence with the conjugacy classes of perfect subgroups of G.

(ii) The number of primitive idempotents of DB(A,G) is equal to the number of

nonconjugate perfect subgroups of G.

(iii) G is solvable if and only if Spec(DB(A,G)) is connected if and only if 0 and

1 are the only idempotents of DB(A,G).

(iv) If |(DB(A,G))∗| = 2, then G is solvable.

Proof : (i) Follows from the above explanation.

(ii) It follows from the bijection between the connected components of the prime

ideal spectrum of a Noetherian commutative ring R and its primitive idempo-

tents.

(iii) By the above explanation we readily have that G is solvable if and only if

Spec(DB(A,G)) is connected. What remains is the content of (ii).

(iv) Suppose G is not solvable. Then Spec(DB(A,G)) is not connected and hence

DB(A,G) has at least one idempotent e different from 0 and 1. Then since the

nontrivial idempotents e and 1 − e are orthogonal and sum up to 1 we have a

decomposition of DB(A,G);


DB(A,G) ' DB(A,G)e×DB(A,G)(1− e).Because both e and 1 − e are different from 0, 1 the rings DB(A,G)e and

DB(A,G)(1− e) are of characteristics 0. Then |(DB(A,G))∗| ≥ 2 and

|(DB(A,G)(1 − e))∗| ≥ 2. Hence |(DB(A,G))∗| ≥ 4 which is a contradiction.

Therefore G is solvable.

�

We know the connected components of Spec(DB(A,G)), but we do not know

much about the equality of two maximal ideals of DB(A,G). We proved that if

I(H, h, P ) and I(K, k,Q) are equal then P = Q and Op(H) =G Op(K) where p is

the characteristic of the field D/P. Before dealing with general case we examine

some special cases including B(G), for instance we will show for abelian groups

G that the prime ideals I(H, h, P ) and I(Op(H), hp′ , P ) of DB(A,G) are equal.

Note that for a group G, any element g of order coprime to p belongs to Op(G)

because gOp(G) has order dividing the order of g ∈ G and |G/Op(G)|. Hence in

particular hp′ ∈ Op(H) for all h ∈ H.


teristic p. Then for any H,K ≤ G;

I(H, 1, P ) = I(K, 1, P ) if and only if Op(H) =G Op(K).

Proof : (⇒) It is 7.18.

(⇐) Suppose Op(H) =G Op(K). Then I(Op(H), 1, P ) = I(Op(K), 1, P ). By 7.13,

I(Op(H), 1, P ) = I(H, 1, P ) and I(Op(K), 1, P ) = I(K, 1, P ).

Thus, I(H, 1, P ) = I(K, 1, P ).

�

If A is taken to be the trivial group then DB(A,G) reduces to B(G). The

results we obtained for DB(A,G) imply all the desired results about the prime

ideals of B(G) which is the content of the next theorem. All facts about the

prime spectrum of the Burnside rings first obtained by Dress in [8].


Let p be a prime number and H ≤ G. Define

I(H, p) = {x ∈ B(G) : SGH(x) ≡ 0 mod(p)},

I(H, 0) = {x ∈ B(G) : SGH(x) = 0}

where SGH is the ring epimorphism from B(G) to Z sending each G−set S to |SH |.Note that SGH = SGH,h|B(G) for any h ∈ H.

Theorem 7.22 Let p be any prime number and H,K be any subgroups of G.

Then

(i) I(H, p) and I(H, 0) are prime ideals of B(G). Moreover, any prime ideal of

B(G) is one of these forms.

(ii) I(H, 0) is a minimal prime ideal, and I(H, p) is a maximal ideal of B(G).

(iii) I(H, 0) ∩ Z = 0, I(H, p) ∩ Z = pZ.(iv) Let m,n be prime numbers (possibly 0). Then

I(H,m) = I(H,n) implies m = n.

(v) I(H, 0) = I(K, 0) if and only if H =G K.

(vi) I(H, p) = I(K, p) if and only if Op(H) =G Op(K).

(vii) I(H, 0) $ I(H, p), and I(H, 0) ⊆ I(K, p) if and only if I(H, p) = I(K, p).

(viii) Let m,n be prime numbers (possibly 0). Then the prime ideals I(H,m) and

I(H,n) of B(G) are in the same connected component of Spec(B(G)) if and only

if S(H) =G S(K).

(ix) The number of primitive idempotents of B(G) is equal to the number of


(x) G is solvable if and only if Spec(B(G)) is connected if and only if 0 and 1 are

the only idempotents of B(G).

Proof : Let A be the trivial group. Then D = Z and DB(A,G) = B(G).

Moreover I(H, h, P ) and I(H, h, 0) reduce to I(H, p) and I(H, 0), respectively.

Now 7.8 gives (i), (ii), (iii), (iv) and (v); (vii) follows from 7.9; (vi) is 7.21;

(viii), (ix) and (x) follow from 7.20.

�

The following result was obtained by Barker also in [1].


Corollary 7.23 The primitive idempotents of B(G), B(A,G) and DB(A,G) are

all the same.

Proof : Let R ≤ S be a ring extension. A primitive idempotent e of R is an

idempotent of S. If e is not primitive in S then it must be a sum of primitive

idempotents of S. Hence, the number of primitive idempotents of S is more than

the number of primitive idempotents of R, and if they have the same number of

primitive idempotents then their primitive idempotents must be the same. By

7.20 and 7.22, the number of primitive idempotents of DB(A,G) and B(G) are

the same. The result follows because B(G) ≤ B(A,G) ≤ DB(A,G).

�

Lemma 7.24 Let (|A|, |G|) be a p−power (possibly 1), and P be a nonzero prime

ideal of D such that D/P has characteristic p. Then for any (H, h) and (K, k) ∈el(A,G);

I(H, h, P ) = I(K, k, P ) if and only if Op(H) =G Op(K).

Proof : For any T ≤ G, O(T ) is the minimal normal subgroup of T such

that T/O(T ) is an abelian group of exponent dividing the order of A. Hence

any element of T (= T/O(T )) has order dividing both |A| and |G|. Thus, T is a

p−group. Now by 7.11, I(H, h, P ) = I(H, 1, P ) and I(K, k, P ) = I(K, 1, P ). But

then the result follows from 7.21.

�

Corollary 7.25 Let A have p−power order, and P be a nonzero prime ideal of

D such that D/P has characteristic p. Then for any (H, h), (K, k) ∈ el(A,G);

I(H, h, P ) = I(K, k, P ) if and only if Op(H) =G Op(K).

Proof : Since (|A|, |G|) is a p−power, the result follows from 7.24.

�


Corollary 7.26 Let G be a p−group, and P be a nonzero prime ideal of

D such that D/P has characteristic p. Then I(H, h, P ) = I(K, k, P ) for all

(H, h), (K, k) ∈ el(A,G).

Proof : It is immediate from 7.24 because for any subgroup T of G, we have

Op(T ) = 1.

�

Now we want to examine the prime ideals of B(A,G) when it is tensored over

some ring of fractions of D. Hence we begin with writing some basic facts about

the ring of fractions of a commutative ring R with 1 (See [11]). A subset S of R

is called multiplicative if 1 ∈ S, and st ∈ S for all s, t ∈ S. It is called proper if

also 0 /∈ S. Given a proper multiplicative subset S of R we define an equivalence

relation on R× S as follows: (a, s) is equivalent to (b, t) if and only if atu = bsu

for some u ∈ S. The equivalence class containing (a, s) is denoted by a/s, and

the set of equivalence classes , denoted by S−1R, becomes a commutative ring

with 1 (0 = 0/1, 1 = 1/1)with respect to the operations: a/s+ b/t = (at+ bs)/st

and (a/s)(b/t) = (ab)/(st). We have a ring homomorphism ι : R → S−1R given

by ι(r) = r/1 for all r ∈ R. This homomorphism is injective if R is an integral

domain. For an ideal A of S−1R, the ideal ι−1(A) of R is denoted by AC and

called the contraction of A. For an ideal a of R, the ideal {a/s ∈ S−1R : a ∈ a}of S−1R is denoted by aE and called the expansion of a. We have a bijective

correspondence between the prime ideals of S−1R and the prime ideals of R not

intersecting S given by A → AC , aE ← a. Let P be a prime ideal of R. Then

R−P = S is a proper multiplicative subset of R and we write RP for S−1R. The

ring RP is called the localization of R at P, and it is a local ring with unique

maximal ideal PE, and also RP/PE 'the quotient field of the integral domain

R/P. If R is a Dedekind domain and S is a proper multiplicative subset, then

R ≤ S−1R is an integral ring extension and the ring S−1R is also a Dedekind

domain. Now let R = D and P be a nonzero prime ideal of D such that D/P

has characteristic p, we have ZpZ ≤ DP , D ≤ DP , and both are integral ring

extensions.


Theorem 7.27 Let G be a p−group, and P be a nonzero prime ideal of D such

that D/P has characteristic p. Then

(i) The rings DPB(A,G) and ZpZB(A,G) are local.

(ii) 0 and 1 are the only idempotents of the rings DPB(A,G) and ZpZB(A,G).

Proof : (i)∏

(H,h)∈Gel(A,G)DP can be seen as an integral ring extension of the

ring DPB(A,G) because the product of the DP−linear extensions of the maps

SGH,h : DB(A,G) → D, (H, h) ∈G el(A,G), is still injective. Hence any prime

ideal of DPB(A,G) is of the form IP (H, h,QE) = {x ∈ DPB(A,G) : SGH,h(x) ∈QE} where Q is prime ideal of D not intersecting D−P. But there are two possi-

bilities for Q, namely 0 and P. Hence maximal ideals of DPB(A,G) are precisely

IP (H, h, PE) for (H, h) ∈ el(A,G).

Let d1, d2 ∈ D be such that d1 + P = d2 + P in the field D/P. Then d1 − d2 ∈ Pand so d1/1 − d2/1 ∈ PE implying that d1/1 + PE = d2/1 + PE in the field

DP/PE ' D/P.

Note that IP (T, t, PE) is the kernel of the map πPE ◦ SGT,t : DPB(A,G)→ DP →DP/P

E where πPE is the canonical ring epimorphism from DP to DP/PE.

By 7.26 I(H, h, P ) = I(K, k, P ) for all (H, h), (K, k) ∈ el(A,G),

and so from 7.10 we have πP ◦ SGH,h = πP ◦ SGK,k where πP is the canonical ring

epimorphism from D to D/P.

For any x = [AνG/V ] ∈ B(A,G) we have SGH,h(x) + P = SGK,k(x) + P. Then

SGH,h(x)/1 + PE = SGK,k(x)/1 + PE, and so πPE ◦ SGH,h(x) = πPE ◦ SGK,k(x).Therefore IP (H, h, PE) = IP (K, k, PE) for all (H, h), (K, k) ∈ el(A,G).

Consequently DPB(A,G) has only one maximal ideal(=it is local).

ZpZB(A,G) ≤ DPB(A,G) is an integral ring extension and hence the maximal

ideals of ZpZB(A,G) can be obtained from the maximal ideals of DPB(A,G) by

intersecting with ZpZB(A,G). Thus ZpZB(A,G) must be a local ring.

(ii) Follows from (i) and the bijective correspondence between the primitive idem-

potents and the connected components of spectrum.

�

Let π be a set of prime numbers and Z(π) be the ring of fractions of Z with

respect to the proper multiplicative subset Z − ∪p∈πpZ. 7.27 has the following

obvious generalization.


Theorem 7.28 Let G be a π−group. Then Z(π)B(A,G) is a local ring (0 and 1

are the only idempotents of Z(π)B(A,G)).

Theorem 7.29 Let A have p−power order. Then the primitive idempotents of

ZpZB(A,G) and ZpZB(G) are the same.

Proof : Let P be a nonzero prime ideal of D such that D/P has characteristic

p. For (H, h) ∈ el(A,G), let IP (H, h, PE) = {x ∈ DPB(A,G) : SGH,h(x) ∈ PE}and Ip(H, (pZ)E) = {x ∈ ZpZB(G) : SGH(x) ∈ (pZ)E}. From 7.25 it follows that

IP (H, h, PE) = IP (K, k, PE) if and only if Op(H) =G Op(K), and from 7.22(vi)

Ip(H, (pZ)E) = Ip(K, (pZ)E) if and only if Op(H) =G Op(K). Hence spectrums

of the rings DPB(A,G) and ZpZB(G) have the same number of connected com-

ponents, and so they have the same number of primitive idempotents. Since

ZpZB(G) ≤ ZpZB(A,G) ≤ DPB(A,G), the result follows.

�

As a generalization of 7.29 we state the following theorem.

Theorem 7.30 Let π be a set of prime numbers and Z(π) be the ring of fractions

of Z with respect to Z−∪p∈πpZ. Then the rings Z(π)B(G) and Z(π)B(A,G) have

the same primitive idempotents for all π−group A.

As an implication of 7.24 we give the next result.

Theorem 7.31 The primitive idempotents of the rings Z(π)B(A,G) and

Z(π)B(G) are the same where π is the set of primes dividing (|A|, |G|).

For a given prime number p and subgroup H of G, let Hp/O(H) be the Sylow

p−subgroup of H/O(H) and Hp′/O(H) be the complement of Hp/O(H).

We have the following immediate consequences; H/O(H) = Hp/O(H) ×Hp′/O(H), O(H) E Hp E H D Hp′ D O(H), Hp ∩ Hp′ = O(H). Take any

automorphism f of H. Then O(H) = f(O(H)) ≤ f(Hp) (because O(H) is a


characteristic subgroup of H) and so f(Hp)/O(H) is a Sylow p−subgroup of the

abelian group H/O(H). Hence f(Hp) = Hp and similarly f(Hp′) = Hp′ . Therefore

Hp and Hp′ are characteristic subgroups of H.

Lemma 7.32 For a prime number p, let (H, h), (K, k) ∈ el(A,G) satisfy the

following conditions;

(C1) H E K and K/H is a p−group,

(C2) xhx−1 ≡ h mod Hp for all x ∈ K,(C3) h ≡ k mod Kp.

Then we have I(H, h, P ) = I(K, k, P ) for any nonzero prime ideal P of D such

that D/P has characteristic p.

Proof : From (C3); h−1kO(K) has p−power order in K and so we have

I(K,h, P ) = I(K, k, P ) by 7.11. But this means πP ◦ SGK,h = πP ◦ SGK,k from

7.10.

Now take any [AνG/V ] ∈ B(A,G) and compute,

πP ◦ SGH,h([AνG/V ]) =∑

gV⊆G,H≤gV

(gν(h) + P )

=∑

gV⊆G,K≤gV

(gν(h) + P ) +∑

gV⊆G,H≤gV,K�gV

(gν(h) + P )

= πp ◦ SGK,h([AνG/V ]) +∑


(gν(h) + P ).

Thus, using πP ◦ SGK,h = πP ◦ SGK,k we get

πP ◦ SGH,h([AνG/V ])− πP ◦ SGK,k([AνG/V ]) =∑


(gν(h) + P ).

From (C2); (h−1(xhx−1))O(H) has p−power order in H and so ν(h)−1ν(xhx−1)

+P is a p−power root of unity in the field D/P for any ν ∈ Hom(H,A). Since

D/P has characteristic p, it has no nontrivial p−power roots of unity. Thus we

must have ν(h) + P = ν(xhx−1) + P for all x ∈ K and ν ∈ Hom(H,A).

Note that the indices of the last sum range in the K/H−set G/V H − G/V K .

Because K/H is a p−group and K/H−orbits of G/V H −G/V K have nontrivial

sizes, any orbit has size divisible by p. Take a gV ∈ G/V H −G/V K and consider


its orbit orbK/H(gV ) = {kgV : kH ∈ K/H}.If x−1gV ∈ orbK/H(gV ) then we havex−1gν(h) + P = gν(xhx−1) + P = gν(h) + P. Therefore,∑


(gν(h) + P ) =∑

gV ∈G/V H−G/V K

(gν(h) + P )

=∑

gV ∈K/HG/VH−G/V K

|orbK/H(gV )|(gν(h) + P ) = P.

So we proved πP ◦ SGH,h = πP ◦ SGK,k implying that I(H, h, P ) = I(K, k, P ).

�

Remark 7.33 Let G be an abelian group, and P be a nonzero prime ideal of D

such that D/P has characteristic p. Then I(H, h, P ) = I(Op(H), hp′ , P ) for all

(H, h) ∈ el(A,G).

Proof : It suffices to note that (Op(H), hp′) and (H, h) satisfy the conditions

in 7.32. (C1) holds trivially , and the condition (C2) disappears because G is

abelian.

Since H/O(H) = Hp/O(H)×Hp′/O(H), any element hO(H) of H can be writ-

ten uniquely as hO(H) = (h1O(H))(h2O(H)) where h1O(H) ∈ Hp/O(H) and

h2O(H) ∈ Hp′/O(H). Since h1O(H) has p−power order, h2O(H) has order not

divisible by p and hO(H) = (h1O(H))(h2O(H)) = (h2O(H))(h1O(H)) we must

have h1O(H) = (hO(H))p = hpO(H) and h2O(H) = (hO(H))p′ = hp′O(H).

Thus (hp′)−1hO(H) = hpO(H) = h1O(H) ∈ Hp/O(H) implying that hp′ ≡ h

mod Hp. Therefore the condition (C3) holds.

�

Let p be a prime number and H be a subgroup of G. Then (Op(H))p′ is a

characteristic subgroup of Op(H) of p−power index. Therefore, (Op(H))p′ is a

normal subgroup of H having p−power index. Then by the minimality of Op(H)

we must have Op(H) = (Op(H))p′ . Thus, Op(H)/O(Op(H)) is a p′−group.


Lemma 7.34 Let H ≤ G be such that H (= H/O(H)) is a p′−group, and P be

a nonzero prime ideal of D such that D/P has characteristic p. Let |A| = pαm

where (p,m) = 1. Then;

(i) The field D/P contains a primitive mth root of unity. In fact, ζpα

+ P is a

primitive mth root of unity in D/P where A =< ζ > .

(ii) The group algebra (D/P )H is semisimple.

(iii) For any ν ∈ Hom(H,A) define a (D/P )−algebra map ev(ν) : (D/P )H →(D/P ) given by ev(ν)(

∑h∈H λhh) =

∑h∈H λhν(h) for all

∑h∈H λhh ∈ (D/P )H.

The maps ev(ν), where ν ranges in Hom(H,A), are all distinct.

(iv) The product map∏

ν∈Hom(H,A) ev(ν) : (D/P )H →∏

ν∈Hom(H,A)(D/P ) is

injective.

Proof : (i) The following three facts can be found in [12](page 57, 58).

(a) Let θ be a (pβ)th primitive root of unity. Then the principal ideal (1− θ)Z[θ]

is the unique prime ideal of Z[θ] lying over pZ.

(b) Let θ be a primitive mth root of unity. If m is divisible by at least two different

prime numbers, then 1− θ is unit in Z[θ].

(c) Let θ be a primitive mth root of unity, and q be a prime number not dividing

m. Then for any prime ideal Q of Z[θ] lying over qZ, 1 − θk ∈ Q if and only if

1− θk = 0. So θ +Q is a primitive mth root of unity in (Z[θ])/Q.

We are assuming that ζ =a primitive nth root of unity, A =< ζ >, D = Z[ζ], and

A ≤ D∗.

(A) Any prime ideal P of D lying over pZ contains 1− ζm :

It is clear that ζm is a primitive (pα)th root of unity. So by the above fact (a),

(1 − ζm)Z[ζm] is the unique prime ideal of Z[ζm] lying over pZ. Now P ∩ Z[ζm]

is a prime ideal of Z[ζm] lying over pZ because Z[ζm] ≤ D. Hence, P ∩ Z[ζm] =

(1− ζm)Z[ζm] implying that 1− ζm ∈ P.(B) Let P be a prime ideal of D lying over pZ. If 0 6= 1− ζk ∈ P, then m divides

k :

1− ζk ∈ P ∩ Z[ζk] which is a prime ideal of Z[ζk] because Z[ζk] ≤ D. Let ζk has

order t. Then since 1− ζk is in a prime ideal of Z[ζk], 1− ζk is nonunit. Thus by

the above fact (b), t cannot be divisible by more than one prime number. Also

from the above fact (c) p must divide t, otherwise 1 − ζk = 0. Hence t = pa for


some natural number a. Now the order of ζk is given by the formula; pa = t =

pαm/(k, pαm). Hence k must divide m.

(C) Let P be a prime ideal of D lying over pZ. The smallest integer k such that

1− ζk ∈ P is m :

Follows from (A) and (B).

Now we know from (C) that 1 − ζm ∈ P and m is the smallest such number.

Then (ζ + P )m = 1 + P , and so ζ + P is an mth root of unity in D/P. On the

other hand the minimality of m implies that ζ + P has order m. Thus ζ + P is a

primitive mth root of unity in D/P.

(ii) D/P is a field of characteristic not dividing H. Thus (D/P )H is semisimple

by Mascke’s Theorem.

(iii) Any element h of H has order dividing m. So for any ν ∈ Hom(H,A), ν(h)

is an mth root of unity in < ζ >≤ D∗. Thus ν(h) belongs to the unique subgroup

of A of order m, namely < ζpα> . Note that ζp

αis a primitive mth root of

unity. Since D/P contains a primitive mth root of unity 1 + ζ we have a group

monomorphism < ζpα>'< ζ + P >≤ (D/P )∗ given by (ζp

α)r 7→ ζr + P for any

integer r. Therefore, the maps ev(ν) are all distinct for all ν ∈ Hom(H,A).

(iv) Since (D/P )H is a semisimple algebra over the field (D/P ) containing all ath

roots of unity for all a dividing the order of H, (D/P )H must be isomorphic to the

direct sum of |H| many (D/P ) by Wedderburn Theorem. Hence there are exactly

|H| = |Hom(H,A)| many distinct (D/P )−algebra maps from (D/P )H to D/P.

However we know that the (D/P )−algebra maps ev(ν), ν ∈ Hom(H,A), are all

distinct. Thus they are precisely all the (D/P )−algebra maps. Consequently,

the product map is injective.

�

Recall that N(H) acts on H by conjugation where N(H) denotes NG(H)/H.

Lemma 7.35 Let H ≤ G be such that H is a p′−group, and P be a nonzero

prime ideal of D such that D/P has characteristic p. Given h ∈ H;

[AνG/H] ∈ I(H, h, P ) for all ν ∈ Hom(H,A) if and only if there is a gH ∈ N(H)

of order p such that gh = h.

Proof : Let x =∑

gH⊆NG(H)g−1h ∈ (D/P )H. Take any ν ∈ Hom(H,A) and


compute;

ev(ν)(x) =∑

gH⊆NG(H)

ν(g−1

h) + P =∑

gH⊆NG(H)

gν(h) + P

= πP ◦ SGH,h([AνG/H]) = 0

Since it is true for all ν ∈ Hom(H,A), x is in the kernel of the product map∏ν∈Hom(H,A) ev(ν) : (D/P )H → (D/P ). By the injectivity of the product map

we must have,

0 = x =∑

gH⊆NG(H)

g−1

h = |stabN(H)(h)|∑

k=N(H)h

k.

In the group algebra (D/P )H, the different elements of H are linearly indepen-

dent over (D/P ). Thus, x = 0 if and only if the characteristic of the field (D/P )

divides |stabN(H)(h)|. Hence, stabN(H)(h) has an element of order p if and only if

[AνG/H] ∈ I(H, h, P ) for all ν ∈ Hom(H,A).

�

Lemma 7.36 Let H ≤ G be such that H is a p′−group, and P be a nonzero

prime ideal of D such that D/P has characteristic p. Let h1, h2 ∈ H. Suppose

that πP ◦ SGH,h1([AνG/H]) = πP ◦ SGH,h2

([AνG/H]) for all ν ∈ Hom(H,A), but

πP ◦ SGH,h1([AνG/H]) 6= 0 for at least one ν ∈ Hom(H,A). Then there is a

gH ∈ N(H) such that gh1 = h2.

Proof : Let x =∑

gH⊆NG(H)g−1h1 and y =

∑gH⊆NG(H)

g−1h2. For any ν ∈

Hom(H,A) we evaluate the image of x − y ∈ (D/P )H under the map ev(ν) :

(D/P )H → D/P ;

ev(ν)(x− y) =∑

gH⊆NG(H)

(ν(g−1

h1)− ν(g−1

h2)) + P

=∑

gH⊆NG(H)

(gν(h1) + P )−∑

gH⊆NG(H)

(gν(h2) + P )

= πP ◦ SGH,h1([AνG/H])− πP ◦ SGH,h2

([AνG/H]) = 0.


Since it is true for all ν ∈ Hom(H,A), x − y is in the kernel of the product

map∏

ν∈Hom(H,A) ev(ν) which is an injective map. Hence x− y = 0 in the group

algebra (D/P )H. Since N(H) acts on H by conjugation we can write x − y as

follows;

0 = x− y = |stabN(H)(h1)|∑

k1=N(H)h1

k1 − |stabN(H)(h2)|∑

k2=N(H)h2

k2.

By 7.35, p cannot divide the orders of N(H)−stabilizers of h1 and h2 because

πP ◦SGH,h1([AνG/H]) is nonzero for some ν ∈ Hom(H,A). Since x−y = 0, by the

linear independency of different elements of H over (D/P ) we can conclude that

k1 = k2 for some k1 =N(H) h1 and k2 =N(H) h2. Thus h1 =N(H) h2, as desired.

�

Lemma 7.37 Let P be a nonzero prime ideal of D such that D/P has char-

acteristic p, and H ≤ G be such that H is a p′−group For h1, h2 ∈ H if

I(H, h1, P ) = I(H, h2, P ), then h1 =N(H) h2 or [AνG/H] ∈ I(H, h1, P ) for all

ν ∈ Hom(H,A).

Proof : Suppose I(H, h1, P ) = I(H, h2, P ). Then by 7.10 πP ◦SGH,h1= πP ◦SGH,h2

.

In particular, πP ◦SGH,h1([AνG/H]) = πP ◦SGH,h2

([AνG/H]) for all ν ∈ Hom(H,A).

Now if [AνG/H] /∈ I(H, h1, P ) for some ν ∈ Hom(H,A) then by 7.36 we must

have h1 =N(H) h2 which completes the proof.

�

Corollary 7.38 Let P be a nonzero prime ideal of D such that D/P has char-

acteristic p, and G be a p′−group. Then for any (H, h), (K, k) ∈ el(A,G);

I(H, h, P ) = I(K, k, P ) if and only if (H, h) =G (K, k).

Proof : (⇒) Suppose I(H, h, P ) = I(K, k, P ). We know from 7.18 that

I(H, h, P ) = I(K, k, P ) implies Op(H) =G Op(K). Since G is a p′−group,

Op(T ) = T for any T ≤ G. Hence H =G K. Suppose H = gK. Then I(H, h, P ) =

I(gK,h, P ) = I(K, k, P ) = I(gK, gk, P ) implying that I(H, h, P ) = I(H, gk, P ).


Since H is a p′−group, from 7.37 we get h =N(H)gk or [AνG/H] ∈ I(H, h, P )

for all ν ∈ Hom(H,A). However, according to 7.35 [AνG/H] ∈ I(H, h, P ) for

all ν ∈ Hom(H,A) if and only if stabN(H)(h) ≤ G contains an element of or-

der p. Since G is a p′−group, stabN(H)(h) has no element of order p. Therefore,

h =N(H)gk. So xh = gk for some x ∈ NG(H). Then x(H, h) = (xH, xh) =

(H, xh) = (gK, gk) = g(K, k). That is, (H, h) =G (K, k).

(⇐) Trivially holds.

�

Theorem 7.39 Let G be a p′−group, and P be a nonzero prime ideal of D such

that D/P has characteristic p. Then the rings DPB(A,G) and CB(A,G) have

the same primitive idempotents.

Proof : By the integrality of the ring extensionDPB(A,G) ≤∏

(H,h)∈Gel(A,G)DP ,

it is clear that the prime ideals of DPB(A,G) are the kernels of the maps SGH,h :

DPB(A,G) → DP and πPE ◦ SGH,h : DPB(A,G) → DP → DP/PE. We denote

these kernels by IP (H, h, 0) and IP (H, h, PE), respectively. The maximal ideals of

DPB(A,G) are of the form IP (H, h, PE) where (H, h) ∈ el(A,G). By 7.38 we con-

clude that IP (H, h, PE) ∼ IP (K, k, PE) (equivalently, I(H, h, P ) = I(K, k, P ))

if and only if (H, h) =G (K, k). Hence, the number of primitive idempotents of

DPB(A,G) is equal to the number of nonconjugate subelements (H, h) ∈ el(A,G)

which is equal to the number of primitive idempotents of CB(A,G). The result

follows, because DPB(A,G) ≤ CB(A,G).

�

The previous theorem implies the following result.

Theorem 7.40 The primitive idempotents of ZpZB(A,G) and QB(A,G) are the

same for any prime number p not dividing the order of G.

Proof : Let P be a nonzero prime ideal of D such that D/P has characteristic p

where p is a prime number not dividing the order of G. From 7.39 we know that

the primitive idempotents of DPB(A,G) are precisely the primitive idempotents


of CB(A,G). Since Q∩DP = ZpZ, QB(A,G)∩DPB(A,G) = ZpZB(A,G) which

completes the proof.

�

Next we find the prime ideals of B(A,G) and their properties by using the

facts we obtained so far. We use the integrality of the ring extensions B(A,G) ≤DB(A,G) and B(A,G) ≤

∏(H,h)∈Gel(A,G)D. We already found almost all facts

appearing in the next theorem. For any (H, h) ∈ el(A,G), and nonzero prime

ideal P of D define

J(H, h, 0) = {x ∈ B(A,G) : SGH,h(x) = 0},

J(H, h, P ) = {x ∈ B(A,G) : SGH,h(x) ∈ P}.

Let G be the Galois group of the extension Q ≤ Q(ζ). Then G acts transitively on

the prime ideals of D lying over pZ where p is any prime number. Restricting any

element σ ∈ G to A we get an automorphism of A. Thus by function composition

G acts on Hom(H,A) ' H for any H ≤ G. Note that G acts also on DB(A,G)

as follows;

(σ,∑

dV,ν [AνG/V ]) 7→ σ(∑

dV,ν [AνG/V ]) =∑

σ(dV,ν)[AνG/V ].

For any prime ideal (possibly 0) A of D, σ(I(H, h,A)) = I(H, σ(h), σ(A)).

Theorem 7.41 Let P,Q be any nonzero prime ideals of D, and (H, h), (K, k) be

any elements of el(A,G). Then

(i) J(H, h, 0) and J(H, h, P ) are prime ideals of B(A,G). Moreover, any prime

ideal of B(A,G) is one of these forms.

(ii) J(H, h, 0) is a minimal prime ideal of B(A,G), and J(H, h, P ) is a maximal

ideal of B(A,G).

(iii) J(H, h, 0) ∩ Z = 0, and J(H, h, P ) ∩ Z = P ∩ Z = pZ where p is the

characteristic of the field D/P.

(iv) J(H, h, P ) = J(K, k,Q) implies that P ∩ Z = Q ∩ Z.(v) Let A,B be prime ideals (possibly 0) of D. Then J(H, h,A) ∼ J(K, k,B) if

and only if S(H) =G S(K).


(vi) The number of primitive idempotents of B(A,G) is equal to the number of


(vii) G is solvable if and only if Spec(B(A,G)) is connected if and only if 0 and

1 are the only idempotents of B(A,G).

Proof : By the integrality of the ring extension B(A,G) ≤ DB(A,G), any

prime ideal of B(A,G) is the intersection of a prime ideal of DB(A,G) with

B(A,G). Note that I(H, h, 0)∩B(A,G) = J(H, h, 0) and I(H, h, P )∩B(A,G) =

J(H, h, P ). Then (i) and (ii) follows. (iii) and (iv) are obvious. For the

rest; note that D is G−stable and (DB(A,G))G = B(A,G). Consequently,

I(H, h,A) ∩ B(A,G) = I(K, k,B) ∩ B(A,G) if and only if there is a σ ∈ Gsuch that σ(I(H, h,A)) = I(H, σ(h), σ(A)) = I(K, k,B). Therefore, J(H, h, P ) =

J(K, k,Q) still implies that Op(H) =G Op(K), and so the results follow.

�

As remarked at the beginning of this chapter some results we obtained so far

can be extended to RB(A,G) where R is not so specific as D. In the following

three remarks we collect some of the results which we obtained for DB(A,G) and

state them for RB(A,G).

Let R be an integral domain of characteristic 0 such that R contains a primitive

nth root of unity to ensure that we have a fixed embedding of A into the unit

group of R. As before, for any (H, h) ∈ el(A,G) we have a ring epimorphism

SGH,h : RB(A,G)→ R where [AνG/V ] 7→∑

gV⊆G,H≤gV

gν(h).

The product map

ψ =∏

(H,h)∈Gel(A,G)

SGH,h : RB(A,G)→∏

(H,h)∈Gel(A,G)

R

is still injective. For any nonzero prime ideal P of R we again define

I(H, h, 0) = {x ∈ RB(A,G) : SGH,h(x) = 0},

I(H, h, P ) = {x ∈ RB(A,G) : SGH,h(x) ∈ P}.

Let also πP : R→ R/P be the canonical ring epimorphism.


Remark 7.42 Let R be an integral domain of characteristic 0 such that R con-

tains a primitive nth root of unity to ensure that A ≤ R∗. Then;

(i) Any prime ideal of RB(A,G) is of the form I(H, h, P ) for some prime ideal P

of R and element (H, h) of el(A,G). Conversely, any I(H, h, P ) is a prime ideal

of RB(A,G)

(ii) I(H, h, 0) is a minimal prime ideal of RB(A,G)

(iii) P is a maximal ideal of R if and only if I(H, h, P ) is a maximal ideal of

RB(A,G)

(iv) For any prime ideal P of R, I(H, h, P ) ∩R = P

(v) For any nonzero prime ideal P of R, I(H, h, 0) $ I(H, h, P )

(vi) I(H, h, P ) = I(K, k,Q) implies P = Q

(vii) I(H, h, 0) = I(K, k, 0) if and only if (H, h) =G (K, k)

(viii) Suppose P is a maximal ideal of R. Then I(H, h, 0) ⊆ I(K, k, P ) if and

only if I(H, h, P ) = I(K, k, P )

(ix) I(H, h, P ) = I(K, k, P ) if and only if πP ◦ SGH,h = πP ◦ SGK,k(x) Let P be a prime ideal of R such that R/P has characteristic p. Then for any

(H, h1), (H, h2) ∈ el(A,G) with h−11 h2O(H) ∈ H is of p−power order, we have

I(H, h1, P ) = I(H, h2, P ).

Proof : Since both rings RB(A,G) and∏

(H,h)∈Gel(A,G)R are finite over R, the

ring extension RB(A,G) ≤∏

(H,h)∈Gel(A,G)R is an integral ring extension. Hence,

all parts of the remark follow by slight modifications of the proofs that we gave

for 7.8, 7.9, 7.10 and 7.11 (indeed, word by word).

�

To study the prime spectrum of RB(A,G) we assume R is Noetherian implying

RB(A,G) is Noetherian.

Remark 7.43 Let R be a Noetherian integral domain of characteristic 0 such

that R contains a primitive nth root of unity to ensure that A ≤ R∗.Then;

(i) Let P be a prime ideal of R and (H, h) ∈ el(A,G). If no prime divisor of |H|is invertible in R, then the prime ideals I(H, h, P ) and I(H, 1, 0) of RB(A,G)

are in the same connected component of Spec(RB(A,G))


(ii) Let P be a prime ideal of R and (H, h) ∈ el(A,G). If no prime divisor of |H|is invertible in R, then the prime ideals I(H, h, P ) and I(S(H), 1, 0) of RB(A,G)

are in the same connected component of Spec(RB(A,G))

(iii) Let P and Q be prime ideals of R, and (H, h), (K, k) ∈ el(A,G) Suppose no

prime divisor of |H||K| is invertible in R. Then S(H) =G S(K) implies that the

prime ideals I(H, h, P ) and I(K, k,Q) of RB(A,G) are in the same connected

component of Spec(RB(A,G)).

Proof : It is clear that the proofs given for 7.12, 7.13, 7.14 and 7.15 still work

under the assumptions given in this remark.

�


that R contains a primitive nth root of unity to ensure that A ≤ R∗. Suppose

further that any nonzero prime ideal of R is maximal, and no prime divisor of

|G| is invertible in R. Then for any prime ideals P and Q of R, the prime ideals

I(H, h, P ) and I(K, k,Q) of RB(A,G) are in the same connected component of

Spec(RB(A,G)) if and only if S(H) =G S(K).

Proof : Under the hypothesis of this remark, evidently 7.19 is still true for

RB(A,G). Hence the result follows from 7.19 and 7.43.

�

We finish generalizations of our results obtained for DB(A,G) after giving a

generalization of one of the remarkable results of Dress appearing in [9].


that R contains a primitive nth root of unity to ensure that A ≤ R∗. Suppose

further that any nonzero prime ideal of R is maximal, and no prime divisor

of |G| is invertible in R. Then G is solvable if and only if Spec(RB(A,G)) is

connected.


Proof : It is immediate from 7.44.

�

Now we collect the results we proved related to idempotents in the following

corollary. The first three parts of the following corollary are already known ((ii)

was obtained by Barker in [1], and (i) and (iii) were obtained by Dress in [9]).

We also achieved to get these three known parts by our slightly different way.

Corollary 7.46 Let A =< ζ >, ζ =a primitive nth root of unity, π =a set

of prime numbers, Z(π) = {a/b ∈ Q : b /∈ ∪p∈πpZ}, and for π = {p} write

Z(π) = ZpZ. Then;

(i) G is solvable if and only if 0 and 1 are the only idempotents of B(A,G)

(ii) The primitive idempotents of B(A,G) and B(G) are the same

(iii) If G is a π−group, 0 and 1 are the only idempotents of Z(π)B(A,G)

(iv) If A is a π−group, the primitive idempotents of Z(π)B(A,G) and Z(π)B(G)

are the same

(v) If π is the set of primes dividing (|A|, |G|), then the primitive idempotents of

Z(π)B(A,G) and Z(π)B(G) are the same

(vi) For any prime number p not dividing |G|, the primitive idempotents of

ZpZB(A,G) and QB(A,G) are the same

(vii) If G is π′−group, then the primitive idempotents of Z(π)B(A,G) and

QB(A,G) are the same.

Proof : All of them except (vii) are obtained so far. However (vii) is trivial

because it is an obvious generalization of (vi).

�

We will close this chapter after finding the primitive idempotents of

Z(π)B(A,G) where π is any set of prime numbers, and G is a nilpotent group.

We will show that there is a bijection between the primitive idempotents of


Z(π)B(A,G) and the primitive idempotents of QB(A,K) where G is a nilpo-

tent group, and K is the unique Hall π′−subgroup of G. I discovered this result,

but the proof I suggested was not complete. A complete proof was given by my

supervisor L. Barker.

For a K−algebra R, a K−algebra homomorphism from R to K is called species.

We use the notation Ipot(R) to denote the set of the primitive idempotents of R.

We begin with giving a slight generalization of 7.46.

Remark 7.47 Let R be an integral domain with characterisric 0, and π be a set

of prime numbers that are not invertible in R. Then;

(i) If G is a π−group, then Ipot(RB(A,G)) = Ipot(B(G))

(ii) If G is a π′−group, then Ipot(RB(A,G)) = Ipot(KB(A,G)) where K is the

field of fractions of R.

Proof : Both part follow from obvious generalizations of 7.46 (iii), (v) and (vii).

�

The next lemma and its proof was given by L. Barker.

Lemma 7.48 Let R be an integral domain with characteristic 0. Let K be the

field of fractions of R. Let A and B be R−algebras, finitely generated and free

as R−modules. Suppose that KA and KB are direct sums of copies of K (Here,

KA = K ⊗R A and KB = K ⊗R B). Given primitive idempotents e and f of A

and B, respectively, then e⊗ f is a primitive idempotent of A⊗R B.

Proof : By replacing A and B with eAe and fBf, we may assume that that

e = 1A and f = 1B. Write e =∑

i∈I ei and f =∑

j∈J fj as the sums of the

primitive idempotents of KA and KB. Thus

KA = ⊕i∈IKei, KB = ⊕j∈JKfj.

Consider a nonzero idempotent ε of A⊗R B. We must show that ε = 1. We have

ε =∑

(i,j)∈T

ei ⊗ fj


for some subset T of I × J. We are to show that T = I × J.Fix an element i ∈ I, let J(i) = {j ∈ J : (i, j) ∈ T}, and suppose that J(i) is

nonempty. We claim that J(i) = J. Let si be the species KA → K such that

si(ei) = 1. Then si(ei′) = 0 for i′ ∈ I−{i}. Since si(1A) = 1 and si(R1A) = R, we

have R ⊆ si(A). But si(A) is finitely generated as an R−module, so R = si(A).

Therefore, we have an R−algebra map

si ⊗ idB : A⊗R B → B, a⊗ b 7→ si(a)b.

The element

si ⊗ idB(ε) =∑j∈J(i)

fj

is an idempotent of B, nonzero because J(i) is nonempty. But 1B is a primitive

idempotent of B, so J(i) = J, as claimed.

Now fix an element j ∈ J, and let I(j) = {k ∈ I : (k, j) ∈ T}. By what we have

shown, i ∈ I(j). In particular, I(j) is nonempty. Interchanging A and B, the

claim established above implies that I(j) = I. Therefore T = I × J, as required.

�

Let H and K be two groups of coprime order. Then any subgroup of H × Kis of the form T × L where T and L are subgroups of H and K, respectively.

For any ν ∈ Hom(T × L,A), define ν1 ∈ Hom(T,A) and ν2 ∈ Hom(L,A) as

ν1(t) = ν((t, 1)) and ν2(l) = ν((1, l)) for all t ∈ T and l ∈ L. So we have a

map Hom(T × L,A) → Hom(T,A) ×Hom(L,A), given by ν 7→ (ν1, ν2), which

is a group isomorphism. This map induces the ring homomorphism given in the

following lemma.

Lemma 7.49 Let H and K be two groups of coprime order. Then the map

B(A,H ×K)→ B(A,H)×B(A,K) given by [AνH×KT×L ] 7→ ([Aν1H/T ], [Aν2K/L])

for any (T × L, ν) ∈ ch(H ×K,A) is a unital ring epimorphism.

Proof : Straightforward checking.

�


By considering dimensions of B(A,H ×K) and B(A,H) × B(A,K) over Z, it

is clear that the map in the previous lemma is not injective in general. To make

it injective, an obvious way is the taking the tensor product of B(A,H) and

B(A,K) over Z instead of taking the direct product of them.

Lemma 7.50 Let H and K be two groups of coprime order. Then the map

ψ : B(A,H × K) → B(A,H) ⊗Z B(A,K) given by ψ([AνH×KT×L ]) = [Aν1H/T ] ⊗

[Aν2K/L]) for any (T × L, ν) ∈ ch(H ×K,A) is a ring isomorphism.

Proof : It is clear.

�

Note that for any integral domain R with characteristic 0, the R−linear exten-

sion of the map ψ given in 7.50 is also a ring isomorphism from RB(A,H ×K)

to RB(A,H)⊗R RB(A,K).

The following theorem was obtained by L. Barker, and its proof below was sug-

gested by him.

Theorem 7.51 Suppose G = P ×Q where P is a π−group and Q is a π′−group,

and π is a set of prime numbers. Also suppose that O is a ring extension of Zwhose field of fractions K is a Galois extension of Q. Let R = O(π) = {x/y ∈ K :

y /∈ ∪p∈πpZ}. Then there is a bijection

Ipot(RB(A,G)) = Ipot(B(P ))× Ipot(KB(A,Q))

such that ε ↔ e ⊗ f provided ε = e ⊗ f. Here, we make the identification

RB(A,G) = RB(A,P ) ⊗R RB(A,Q) given by the R−linear extension of the

map ψ introduced in 7.50.

Proof : First suppose that R has enough roots of unity. By 7.48, there is a

bijection

Ipot(RB(A,G))↔ Ipot(RB(A,P ))× Ipot(RB(A,Q))

described by tensor products, e⊗f ↔ (e, f). The required conclusion now follows

from 7.47 in this case.


For the general case, let K ≤ K′ be a Galois extension such that K′ has enough

roots of unity. Let O′ be the ring of algebraic integers in K′, and let R′ =

O′(π). The Galois group Γ of the extension K ≤ K′ acts on R′, and the Γ−fixed

subalgebra of R′ is R. Letting Γ act in the evident way on R′B(A,G), then the

Γ−fixed subalgebra is RB(A,G). So

Ipot(RB(A,G)) = Ipot(R′B(A,G))Γ.

Similarly, we have

Ipot(KB(A,Q)) = Ipot(K′B(A,Q))Γ.

On the other hand, the bijection

Ipot(R′B(A,G))↔ Ipot(B(P ))× Ipot(K′B(A,G))

is invariant under Galois automorphisms. Therefore

Ipot(RB(A,G)) = Ipot(R′B(A,G))Γ

↔ (Ipot(B(P )))× Ipot((K′B(A,Q))Γ)

= Ipot((B(P ))Γ)× Ipot((K′B(A,Q))Γ)

= Ipot(B(P ))× Ipot(KB(A,Q)).

�

Theorem 7.51 implies the following result which is about the primitive idem-

potents of Z(π)B(A,G) where G is a nilpotent group and π is any set of prime

numbers.

Theorem 7.52 Let π be a set of prime numbers and Z(π) = {a/b ∈ Q : b /∈∪p∈πpZ}. For a nilpotent group G, there is a bijective correspondence between the

primitive idempotents of Z(π)B(A,G) and the primitive idempotents of QB(A,Q)

where Q is the unique Hall π′−subgroup of G.

Proof : Sine G is nilpotent, G = P ×Q where P is the unique Hall π−subgroup

of G, and Q is the unique Hall π′−subgroup of G. Let O = Z. Then R = Z(π)

and K = Q. So by theorem 7.51, there is a bijection

Ipot(Z(π)B(A,G)) = Ipot(B(P ))× Ipot(QB(A,Q)).


Then the result follows because from 7.47, Ipot(B(P )) = Ipot(Z(π)B(A,P )) =

{1} (by 7.46, Z(π)B(A,P ) is a local ring).

�

By the previous theorem, we know the primitive idempotents of Z(π)B(A,G)

where G is a nilpotent group and π is any set of prime numbers, because for any

group H the primitive idempotents of QB(A,H) are known, and obtained by

Barker in [1].

Chapter 8

Some Further Maps

In this chapter, we collect together some miscellaneous further material on maps

between monomial Burnside rings.

8.1 B(G)→ B(A,G)

We find the images of the primitive idempotents of CB(G) under the map ψ1

defined in 3.4. We repeat its definition below.

For a G−set S, let AS = {as : a ∈ A, s ∈ S} be the set of formal products.

Thus, a1s1 = a2s2 if and only if s1 = s2 and a1 = a2. We let AG act on AS as:

(bg)(as) = (ab)(gs) for all bg ∈ AG and s ∈ S. Then, AS becomes an A−fibred

G−set and we have a well-defined map

ψ1 : B(G) → B(A,G) given by [S] 7→ [AS] for any G−set S. In 3.4 we proved

that ψ1 is a unital ring monomorphism and ψ1([G/V ]) = [AτG/V ] for any V ≤ G

where τ is the trivial group homomorphism V → A. In the following remark we

consider the C−linear extension of ψ1 for which we still use the same notation.

Remark 8.1 (i) For any G−set S and (H, h) ∈ el(A,G) we have

SGH,h(ψ1([S])) = SGH([S]).

83

CHAPTER 8. SOME FURTHER MAPS 84

(ii) For any primitive idempotent eGH of CB(G);

ψ1(eGH) =

∑h

eGH,h

where h runs over all distinct representatives of N(H)−orbits of the N(H)−set

H/O(H).

Proof : (i) It suffices to prove the desired result for transitive G−sets. Take

S = G/V. Then using ψ1([G/V ]) = [AτG/V ];

SGH,h(ψ1([G/V ])) = SGH,h([AτG/V ]) =∑

gV⊆G,H≤gV

1 = SGH([G/V ]).

(ii) For some complex numbers λK,k;

ψ1(eGH) =

∑(K,k)∈Gel(A,G)

λK,keGK,k,

where by part (i)

λK,k = SGK,k(ψ1(eGH)) = SGK(eGH) =

{1, H =G K

0, otherwise.Hence the result follows.

�

8.2 B(A,G)→ B(G)

We find the images of the primitive idempotents of CB(A,G) under the map ψ2

defined in 3.5. Remember that ψ2 is a unital ring epimorphism from B(A,G) to

B(G) given by ψ2([AνG/V ]) = [G/V ] for any (V, ν) ∈ ch(A,G).

Remark 8.2 (i) For any A−fibred G−set S = AX and H ≤ G;

SGH(ψ2([S])) = SGH,1([S]).

(ii) For any primitive idempotent eGH,h of CB(A,G);

ψ2(eGH,h) =

{eGH , h ∈ O(H)

0, otherwise.


Proof : (i) There is no loss in taking S to be transitive. Let S = AνG/V. Then

using ψ2([AνG/V ]) = [G/V ],

SGH(ψ2([AνG/V ])) = SGH([G/V ]) = SGH,1([AνG/V ]).

(ii) For some complex numbers λK ;

ψ2(eGH,h) =

∑K≤GG

λKeGK .

Then using part (i);

λK = SGK(ψ2(eGH,h)) = SGK,1(e

GH,h) =

{1, (H, h) =G (K, 1)

0, otherwise.Hence, the result follows.

�

8.3 B(A,G)→ B(A′, G)

Let A and A′ be two cyclic groups, and f be a group homomorphism from A

to A′. So f is given by f(a) = bn for some natural number n where a and b are

the respective generators of A and A′. Note that f ◦ ν ∈ Hom(V,A′) for any

ν ∈ Hom(V,A). Thus, we can transform the transitive A−fibred G−set AνG/V

to the transitive A′−fibred G−set A′f◦νG/V.

Lemma 8.3 Let (V, ν), (W,ω) ∈ ch(A,G) and g ∈ G. Then we have

(i) g(f ◦ ν) = f ◦ gν ∈ Hom(gV,A′).

(ii) f ◦ (ν.ω) = (f ◦ ν).(f ◦ ω) ∈ Hom(V ∩W,A′).

(iii) If AνG/V 'AG AωG/W, then A′f◦νG/V 'A′G A′

f◦ωG/W.

Proof : (i) For any v ∈ V we compute that

g(f ◦ ν)(gv) = f ◦ ν(v) = f ◦ gν(gv).

(ii) It is clear.

(iii) AνG/V 'AG AωG/W if and only if (V, ν) =G (W,ω). Then (gV, gν) = (W,ω)


for some g ∈ G. But then by part (i);

(W, f ◦ ω) = (gV, f ◦ gν) = (gV, g(f ◦ ν))

implying that A′f◦νG/V 'A′G A′

f◦ωG/W.

�

Now by the above lemma (iii), we have a well-defined map

ψ : B(A,G) → B(A′, G) given by ψ([AνG/V ]) = [A′f◦νG/V ] for all (V, ν) ∈

ch(A,G).

Theorem 8.4 (i) ψ is a (unital) ring homomorphism.

(ii) For any A−fibred G−set S = AX and (H, h) ∈ el(A′, G);

A′SGH,h(ψ([S])) = AS

GH,hn([S]).

(iii) For any (K, k) ∈ el(A,G);

ψ(AeGK,k) =

∑hn=NG(K)k

A′eGK,h.

Proof : (i) By using the part (i) and (ii) of the above lemma;

ψ([AνG/V ][AωG/W ]) = ψ(∑

V gW⊆G

[Aν.gωG/V ∩ gW ])

=∑

V gW⊆G

[A′f◦(ν.gω)G/V ∩ gW ]

=∑

V gW⊆G

[A′(f◦ν).g(f◦ω)G/V ∩ gW ]

= [A′f◦νG/V ][A′

f◦ωG/W ] = ψ([AνG/V ])ψ([AωG/W ]).

So ψ is multiplicative and hence a ring homomorphism.

(ii) Since f : A =< a >→ A′ =< b > with f(a) = bn, we have f ◦ ν(v) = ν(vn)

for any (V, ν) in ch(A,G) and v ∈ V. By this observation;

A′SGH,h(ψ([AνG/V ])) = A′S

GH,h([A

′f◦νG/V ])


=∑

gV⊆G,H≤gV

g(f ◦ ν)(h)

=∑

gV⊆G,H≤gV

gν(hn)

= ASGH,hn([AνG/V ]).

(iii) For some elements λH,h of C;

ψ(AeGK,k) =

∑(H,h)∈Gel(A′,G)

λH,hA′eGH,h.

Then by part (ii);

λH,h = A′SGH,h(ψ(Ae

GK,k)) = AS

GH,hn(Ae

GK,k) =

{1, (H, hn) =G (K, k)

0, otherwise.Hence,

ψ(AeGK,k) =

∑(H,h)∈Gel(A′,G),(H,hn)=G(K,k)

A′eGH,h =

∑hn=NG(K)k

A′eGK,h.

�

8.4 B(A,H)→ B(A,G)

Let G and H be finite groups and α : G→ H be a group homomorphism. Then

any H−set S can be viewed as a G−set with the G action on S:

gs 7→ α(g)s.

Hence any A−fibred H−set S can be viewed as an A−fibred G−set with the

same action of A and the above action of G. Let α∗(S) denote this new fibred

set. Of course, as a set α∗(S) = S.

It is clear that if S 'AH T then α∗(S) 'AG α∗(T ). Therefore, we have a well-

defined map

ψ : B(A,H)→ B(A,G) where [S] 7→ [α∗(S)]

for any A−fibred H−set S.


Theorem 8.5 (i) The map ψ is a unital ring homomorphism.

(ii) If α is surjective, then for any V ≤ H and ν ∈ Hom(V,A)

ψ([AνH/V ]) =|α−1(V )||Ker(α)||V |

[Aν◦αG/α−1(V )].

(iii) For any A−fibred H−set S = AX and (W,w) ∈ el(A,G);

SGW,w(ψ([S])) = SHα(W ),α(w)([S]).

(iv) For any primitive idempotent eHV,v of CB(A,H)

ψ(eHV,v) =∑

(W,w)∈Gel(A,G),(α(W ),α(w))=H(V,v)

eGW,w.

Proof : (i) It is clear because ψ(S) = S for any A−fibred H−set.

(ii) Remember [AνH/V ] represents the AH−isomorphism class of the A−free

AH−set AH/4(V,ν) where 4(V,ν) is the subgroup {ν(v−1)v : v ∈ V } of AH.

Find the AG−stabilizer of an element ah4(V,ν) of α∗(AH/4(V,ν)) = AH/4(V,ν):

bk is in the stabilizer if and only if (bk)ah4(V,ν) = ah4(V,ν). But then since

bα(k)h4(V,ν) = h4(V,ν) is equivalent to bh−1α(k)h ∈ 4(V,ν), we have h−1α(k)h =

v ∈ V and b = ν(v−1). So k ∈ α−1(hV ). If α is surjective then there exists an x ∈G such that α(x) = h. Then b = ν(v−1) = ν(h−1α(k−1)h) = ν(α(x−1k−1x)) =xν ◦ α(k−1) and k ∈ xα−1(V ). Hence;

stabAG(ah4(V,ν)) = {xν ◦ α(k−1)k : k ∈ xα−1(V )} = 4(α−1(V ),ν◦α). Therefore we

must have ψ([AνH/V ]) = n[Aν◦αG/α−1(V )] for some natural number n because

the AG−stabilizer of ah4(V,ν) does not depend on ah up to isomorphism of

AG−sets. Moreover n can be determined by counting the sizes of both sides

|AνH/V ]| = n|[Aν◦αG/α−1(V )]|,

|A||H : V | = n|A||G : α−1(V )|.

(iii) Since α∗(AX) = AX for any A−fibred H−set S = AX, we have;

(a) g ∈ stabG(Ax) if and only if gAx = Ax which is to say that α(g)Ax = Ax, or

equivalently α(x) ∈ stabH(Ax). So α(stabG(Ax)) = stabH(Ax) ∩ α(G).

(b) For W ≤ G; W ≤ stabG(Ax) if and only if α(W ) ≤ stabH(Ax).


(c) For any x ∈ X, let ϑGx and ϑHx denote the uniquely determined elements of

Hom(stabG(Ax), A) and Hom(stabH(Ax), A) by the conditions gx = ϑGx (g)x and

hx = ϑHx (h)x. We can compute that for any w ∈ W ≤ stabG(Ax) we have;

ϑGx (w)x = wx = α(w)x = ϑHx (α(w))x implying that ϑGx (w) = ϑHx (α(w)). Now

SGW,w(ψ([S])) = SGW,w(α∗(AX)) =∑

x∈X,W≤stabG(Ax)

ϑGx (w)

=∑

x∈X,α(W )≤stabH(Ax)

ϑHx (α(w)) = SHα(W ),α(w)(AX) = SHα(W ),α(w)([S]).

(iv) Clearly for some complex numbers λW,w we have

ψ(eHV,v) =∑

(W,w)∈Gel(A,G)

λW,weGW,w,

where (by using part (iii))

λW,w = SGW,w(α∗(eHV,v)) = SHα(W ),α(w)(eHV,v).

Then the result follows because SHα(W ),α(w)(eHV,v) takes only two values 0 or 1 (takes

value 1 if and only if (α(W ), α(w)) =H (V, v)).

�

8.5 B(A,G)→ B(G)

Suppose S = AX is an A−fibred G−set. Hence, in particular S is a G−set.

Moreover, for A−fibred G−sets S and T, if S 'AG T then S 'G T. Thus, we

have a well-defined map

ψ : B(A,G) → B(G) given by ψ([S]) = [GS] for any A−fibred G−set S where

the notation GS means that we regard S as a G−set.

Remark 8.6 (i) ψ is a Z−module homomorphism.

(ii) For any (V, ν) ∈ ch(A,G)

ψ([AνG/V ]) =|A|

|V : Kerν|[G/Kerν].

(iii) ψ is not multiplicative, not injective, not surjective.


Proof : (i) Obvious.

(ii) AνG/V = AG/4(V,ν) where 4(V,ν) = {ν(v−1)v : v ∈ V } which is a subgroup

of AG. Put 4(V,ν) = 4.We find the G−stabilizer of an element ag4 of AνG/V ;

h ∈ G is in the stabilizer if and only if ahg4 = ag4 which is equivalent to

g−1hg ∈ 4. By the definition of 4, g−1hg ∈ 4 if and only if g−1hg ∈ V and

ν((g−1hg)−1) = 1 which is to say that h ∈ gV and h ∈ Kergν = g(Kerν), or

equivalently h ∈ g(Kerν). Hence, orbG(ag4) 'G G/Kerν which does not depend

on ag. Therefore AνG/V = nG/Kerν for some natural number n which can be

determined by counting of the elements of both sets.

(iii) Obvious from part (ii).

�

8.6 B(A,G)→ B(A,AG)

Suppose S = AX is an A−fibred G−set. Thus S is an AG−set which is A−free

implying that it is also an A−fibred AG−set. It is obvious that if S and T

are isomorphic A−fibred G−sets, then they are also isomorphic as A−fibred

AG−sets. Hence, we have a well-defined map

ψ : B(A,G)→ B(A,AG) given by ψ([S]) = [S] for any A−fibred G−set S.

Remark 8.7 (i) ψ is a Z−module monomorphism.


ψ([AνG/V ]) = [AνA

AG

AV]

where νA ∈ Hom(AV,A) is given by νA(av) = aν(v) for all av ∈ AV.(iii) ψ is not multiplicative, not surjective.

Proof : (i) Additivity of ψ is clear. It is injective because A(AG)−isomorphism

implies AG−isomorphism.


(ii) AνG/V = AG/4(V,ν) where 4(V,ν) = {ν(v−1)v : v ∈ V } which is a subgroup

of AG. Put 4(V,ν) = 4.It is clear that AνG/V is a transitive A−fibred AG−set and so

AνG/V = orbA(AG)(4) 'A(AG)A(AG)

stabA(AG)(4)

where 4 = 1.14 ∈ AG/4. Now;

a(bg) ∈ A(AG) is in the A(AG)−stabilizer of 4 if and only if a(bg)4 = 4 which

is equivalent to g ∈ V and ν(g−1) = ab. But this holds if and only if bg ∈ AV and

νA((bg)−1) = a, or equivalently a(bg) ∈ {νA((bg)−1)(bg) : bg ∈ AV } = 4(AV,νA).

Hence,

AνG/V 'A(AG) AνAAG/AV which completes the proof.

(iii) It is clear from part (ii).

�

8.7 B(A,G)→ B(A)

Suppose S = AX be an A−fibred G−set. Let S \G be the set of G−orbits of S.

Thus

S \G = {orbG(s) : s ∈ S}.

We let A act on S \ G as: (a, orbG(s)) 7→ a(orbG(s)) = orbG(as). It is clear

that isomorphic A−fibred G−sets have isomorphic (as A−sets ) G−orbit sets.

Therefore, we have well-defined map

ψ : B(A,G)→ B(A) given by ψ([S]) = [S \G] for any A−fibred G−set S.

Remark 8.8 (i) If S is a transitive A−fibred G−set, then S \ G is a transitive

A−set.


ψ([AνG/V ]) = [A/ν(V )].

(iii) ψ is a Z−module homomorphism.

(iv) ψ is not multiplicative, not injective, not surjective.


Proof : (i) Suppose S is a transitive A−fibred G−set. Take any two elements

orbG(s), orbG(s′) from S \ G. By the transitivity of S there is an ag ∈ AG such

that ags = s′ implying that a(orbG(s)) = orbG(s′). Thus, S \ G is a transitive

A−set.

(ii) Since ψ([AνG/V ]) is transitive, we must have ψ([AνG/V ]) = [orbA(orbG(4))]

where 4 = {ν(g−1)g : g ∈ G} ≤ AG and AνG/V = AG/4. Now, a ∈ A is in the

A−stabilizer of orbG(4) if and only if orbG(a4) = orbG(4) which is to say that

ag4 = 4 for some g ∈ G. But ag4 = 4 for some g ∈ G if and only if g ∈ Vand ν(g−1) = a, or equivalently a ∈ ν(V ). Therefore,

ψ([AνG/V ]) = [orbA(orbG(4))] = [A/stabA(orbG(4))] = [A/ν(V )].

(iii) and (iv) They are clear from part (ii).

�

8.8 B(A,G)→ B(AG)

Let ν : V → A be a group homomorphism. Then observing gν(gV ) = ν(V ) we

can see the map

ψ : B(A,G)→ B(AG) given by [AνG/V ] 7→ [ AGν(V )V

] is a well-defined map which

is not multiplicative.

8.9 B(A1 × A2, G)→ B(A1, G)×B(A2, G)

Since our fibre group A is abelian, it is of interest to consider a direct product

decomposition A = A1 ×A2. Let π1 and π2 be the respective projections from A

to A1 and A2. Define a map

ψ : B(A,G) → B(A1, G) × B(A2, G) given for all (V, ν) ∈ ch(A,G) by

ψ([AνG/V ]) = ([A1π1◦νG/V ], [A2π2◦νG/V ]).

Remark 8.9 (i) πi ◦ (ν.µ) = (πi ◦ν).(πi ◦µ) for any H,K ≤ G, ν ∈ Hom(H,A),


and µ ∈ Hom(K,A) where i = 1, 2.

(ii) For any g ∈ G, H ≤ G, and ν ∈ Hom(H,A) we have πi ◦ (gν) = g(πi ◦ ν)where i = 1, 2.

(iii) For any H ≤ G, ν ∈ Hom(H,A) we have Kerν ≤ Ker(πi ◦ ν) and Kerν =

Ker(π1 ◦ ν) ∩Ker(π2 ◦ ν) where i = 1, 2.

(iv) ψ is a unital ring homomorphism.

Proof : (i), (ii) and (iii) follow from easy calculations.

(iv) Using (i) and (ii) it is clear that ψ respects the multiplication of two transitive

A−fibred G−sets. Hence, the result follows.

�

8.10 The Number Of Orbits

Let S = AX be an A−fibred G−set. In particular S is an AG−set, a G−set,

and an A−set. So we have the following three maps:

OAG : B(A,G)→ Z, OAG([S]) =the number of AG−orbits of S

OG : B(A,G)→ Z, OG([S]) =the number of G−orbits of S

OA : B(A,G)→ Z, OA([S]) =the number of A−orbits of S.

Remark 8.10 (i) OAG is a Z−module homomorphism.

(ii) OAG([AX]) = 1|A||G|

∑ag∈AG |AX<ag>|.

(iii) OAG([AνG/V ][AωG/W ]) =the number of double coset representatives of

(V,W ) in G.

Proof : For (ii) we use the result of Burnside which counts the number of orbits

(it is stated at the beginning of chapter 1). The other parts are trivial.

�


Remark 8.11 (i) OG is a Z−module homomorphism.

(ii) OG([AνG/V ]) = |A||V :Kerν| .

Proof : (i) is trivial, and (ii) follows from the result of Burnside which counts

the number of orbits.

�

Remark 8.12 (i) OA is a Z−algebra homomorphism.

(ii) OA([AX]) = |X|.(iii) OA([AνG/V ]) = |G : V |.

Proof : Since any A−fibred G−set is an A−free set, the results are straightfor-

ward.

�

Since OA is multiplicative, it is more important than the first two maps. Some

properties of the map OA is given in the next remark.

Remark 8.13 (i) KerOA = J(G, 1, 0) = {x ∈ B(A,G) : SGG,1(x) = 0}.(ii) If x ∈ B(A,G) is unit in B(A,G), then OA(x) ∈ {−1,+1}.

Proof : (i) KerOA is a prime ideal of B(A,G) which is not maximal. Hence,

we know from chapter 7 that KerOA = J(H, h, 0) for some (H, h) ∈ el(A,G).

However, it is clear that [AνG/V ] − [AµG/V ] ∈ KerOA for all V ≤ G and

ν, µ ∈ Hom(V,A). Thus KerOA must be J(G, 1, 0).

(ii) It is obvious.

�

Chapter 9

The Ring B(A,G)

In this short chapter we study some ring theoretic properties of the monomial

Burnside rings. We are assuming that A is a finite cyclic group regarded as a

subgroup of C∗. As with the previous chapter, this is a compendium of further

results and observations, recorded with a view to subsequent development.

Recall that for any (H, h) ∈ el(A,G), SGH,h : B(A,G) → C is the

ring homomorphism given for any (V, ν) ∈ ch(A,G) by SGH,h([AνG/V ]) =∑gV⊆G,H≤gV

gν(h). Also the injectivity of the product map∏

(H,h)∈Gel(A,G) SGH,h :

B(A,G) →∏

(H,h)∈Gel(A,G) C implies that two elements x and y of B(A,G) are

equal if and only if SGH,h(x) = SGH,h(y) for all (H, h) ∈ el(A,G).

Let R be a unital subring of S. We write RS to imply that we are regarding S

as an R−module. We have Z ≤ B(G) ≤ B(A,G) via the embeddings 1 7→ [G/G],

[G/V ] 7→ [AτG/V ]. So for instance if we write B(G)B(A,G) this means that we

are regarding B(A,G) as a B(G)−module.

Remark 9.1 ZB(G), B(G)B(G), ZB(A,G), B(G)B(A,G), B(A,G)B(A,G) are all

Noetherian modules but not Artinian.

Proof : Since both B(G) and B(A,G) are finite over Z, the results follow because

Z is a Noetherian but not Artinian (as a ring, or equivalently as a module over

95

CHAPTER 9. THE RING B(A,G) 96

itself).

�

Remark 9.2 The only nilpotent element of B(A,G) is 0. So, the nilradical of

the ring B(A,G) is the 0 ideal.

Proof : Let x ∈ B(A,G) be a nilpotent element. Then xn = 0 for some natural

number n. We know from 5.1 that

x =∑

(H,h)∈Gel(A,G)

SGH,h(x)eGH,h.

Then it follows from xn = 0 that

0 =∑

(H,h)∈Gel(A,G)

(SGH,h(x))neGH,h

which implies SGH,h(x) = 0 for all (H, h) ∈ el(A,G) because SGH,h(x) is a complex

number for any (H, h) ∈ el(A,G). Hence, x = 0.

�

Remark 9.3 The Jacobson radical of B(A,G) is the 0 ideal.

Proof : Let R ≤ S be a ring extension of commutative rings. Then from

7.4(Maximality) it is clear that J(R) = R ∩ J(S) where for any ring T, J(T )

denotes the Jacobson radical of the ring T. Let D be as in chapter 7. Using

the integrality of the ring extension DB(A,G) ≤∏

(H,h)∈Gel(A,G)D we conclude

that J(DB(A,G)) = 0 because J(D) = 0. Hence, J(B(A,G)) = 0 because

B(A,G) ≤ DB(A,G) is an integral extension.

�

Remark 9.4 If x ∈ B(A,G) is a zero divisor, then SGH,h(x) = 0 for some

(H, h) ∈ el(A,G). Hence the zero divisors of B(A,G) belong to the union of

the minimal prime ideals of B(A,G).


Proof : Let x be a nonzero zero divisor of B(A,G). Then there is a nonzero y

in B(A,G) such that xy = 0. Now using (see 5.1)

x =∑

(H,h)∈Gel(A,G)

SGH,h(x)eGH,h, y =

∑(H,h)∈Gel(A,G)

SGH,h(y)eGH,h

we see from xy = 0 that SGH,h(x)SGH,h(y) = 0 for all (H, h) ∈ el(A,G). Since y is

nonzero, the result follows.

�

Remark 9.5 Any ring homomorphism ψ : B(A,G)→ C is of the form SGH,h for

some (H, h) ∈ el(A,G).

Proof : The map ψ extends by C−linear extension to a C−algebra map from

CB(A,G) to C. So the result follows from 4.11 for which we referred [1]. The

same result appears also in [9].

�

Remark 9.6 An element x ∈ B(A,G) belongs to B(G) if and only if SGH,h(x) =

SGH,1(x) for all (H, h) ∈ el(A,G).

Proof : See [1].

�

Remark 9.7 Let ψ : B(A,G) → B(A,G) be a ring homomorphism. Then for

any (H, h) ∈ el(A,G) there exists a (K, k) ∈ el(A,G) such that SGH,h ◦ ψ = SGK,k.

Hence any ring endomorphism ψ of B(A,G) induces a map ψ : el(A,G) →el(A,G) given by the condition: SGH,h ◦ ψ = SG

ψ((H,h))for all (H, h) ∈ el(A,G).

Proof : It is obvious since any ring homomorphism from B(A,G) to C is of the

form SGH,h for some (H, h) ∈ el(A,G).

�

Let G\el(A,G) denote the set of G−orbit representatives of the G−set el(A,G).

By the previous remark any ring endomorphism of B(A,G) induces a map from


G \ el(A,G) to G \ el(A,G). However, the converse may not be true since given

any map ψ : G \ el(A,G) → G \ el(A,G) the image of the map ψ is in general

belongs to CB(A,G).

Remark 9.8 Let ψ : G \ el(A,G) → G \ el(A,G) be any map. Define ψ :

CB(A,G) → CB(A,G) by the condition: SGH,h ◦ ψ = SGψ((H,h))

for all (H, h) ∈el(A,G). Then ψ is a ring homomorphism.

Proof : Take any (H, h) ∈ el(A,G). Then for any x, y ∈ B(A,G)

SGH,h(ψ(x+ y)) = SGψ((H,h))

(x+ y) = SGψ((H,h))

(x) + SGψ((H,h))

(y)

= SGH,h(ψ(x)) + SGH,h(ψ(y)) = SGH,h(ψ(x) + ψ(y)).

Since it is true for all (H, h) ∈ el(A,G), ψ(x + y) = ψ(x) + ψ(y). Similar calcu-

lations shows that ψ(xy) = ψ(x)ψ(y).

�

The previous two remarks imply that there are finitely many (at most |G \el(A,G)|) ring endomorphisms of B(A,G).

Theorem 9.9 Let ψ : B(A,G)→ B(A,G) be a ring homomorphism. Then ψ is

injective if and only if it is surjective. In fact, it is almost true also for Z−module

endomorphisms of B(A,G). If ψ : B(A,G)→ B(A,G) is a surjective Z−module

homomorphism, then ψ is injective.

Proof : (⇒) Suppose ψ is injective. Since there are finitely many ring en-

domorphisms of B(A,G), not all of ψ, ψ2, ..., ψn, ... can be distinct. So there

are natural numbers n1, n2 such that ψn1 = ψn2 . To show that ψ is surjec-

tive, take any x ∈ B(A,G). Then ψn1(x) = ψn2(x) implying by the injectiv-

ity of ψ that ψn1−1(x) = ψn2−1(x). Using injectivity of ψ inductively we get

x = ψm(x) = ψ(ψm−1(x)). So ψ is surjective.

(⇐) Suppose ψ is surjective. Since B(A,G) is Noetherian (as both Z−module

and B(A,G)−module), the chain Kerψ ⊆ Kerψ2 ⊆ ...Kerψn ⊆ ... cannot be


infinite. So there is a natural number n such that Kerψn = Kerψn+1. We

show that Kerψn−1 = Kerψn. Let x ∈ Kerψn. Then by the surjectivity of

ψ, there is a y ∈ B(A,G) such that ψ(y) = x. Then ψn+1(y) = ψn(x) and

so y ∈ Kerψn+1 = Kerψn implying that ψn−1(x) = ψn−1(ψ(y)) = ψn(y). So

x ∈ Kerψn−1, and hence Kerψn−1 = Kerψn. Proceeding in this way we can

show that Kerψ2 = Kerψ. To show that ψ is injective, take any z ∈ Kerψ.

Then since ψ is surjective, there is a t ∈ B(A,G) such that z = ψ(t). Now from

ψ(z) = ψ2(t) it follows that t ∈ Kerψ2 = Kerψ and so z = ψ(t) = 0. Thus ψ is

injective.

�

Remark 9.10 B(A,G) has no minimal ideals.

Proof : It is trivial because Z is embeddable in B(A,G).

�

Lastly we show below that we can extend some maps from B(G) to B(A,G).

The following result is immediate from 8.1. We give an alternative proof that does

not require the classification of the species of the monomial Burnside algebras.

Remark 9.11 Let K be an algebraically closed field. Then any ring homomor-

phism ψ : B(G)→ K can be extended to a ring homomorphism ψ : B(A,G)→ K.

Proof : Let ψ : B(G)→ K be a ring homomorphism which is nonzero (otherwise

the result is trivial). Then B(G)/Kerψ is a subring of K, and so it is an integral

domain. Put P = Kerψ which is a prime ideal of B(G). Let S = B(G) − P.

Then S is a proper multiplicative subset of both B(G) and B(A,G). We consider

the ring of fractions of B(G) and B(A,G) with respect to S. For notations and

details about the ring of fractions of commutative rings see chapter 7. So now we

have two new rings B(G)P and S−1B(A,G).

Define ψ1 : B(G)P → K as ψ1(as) = ψ(a)ψ(s)−1 for all a

s∈ B(G)P . It can be


checked easily that ψ1 is a ring homomorphism. Now Kerψ1 is a proper ideal

of B(G)P , and so it is contained in a maximal ideal of B(G)P . However B(G)P

is a local ring with its unique maximal ideal PE. Hence, Kerψ1 ⊆ PE. On the

other hand, if as∈ PE then there is a b ∈ P and t ∈ S such that a

s= b

tand

so ψ1(as) = ψ1(

bt) implying that ψ(b)ψ(t)−1 = ψ(a)ψ(s)−1. But b ∈ P = Kerψ

gives that as∈ Kerψ1 = PE. Hence, PE = Kerψ1. As a result, B(G)P/P

E =

B(G)P/Kerψ1 is a field.

Note that B(G)P ≤ S−1B(A,G) is an integral ring extension. Also Kerψ1 = PE

is a maximal ideal of B(G)P . Then by 7.4(Lying Over) there is a maximal ideal

a of S−1B(A,G) such that Kerψ1 = PE = a ∩B(G)P .

Define φ : B(G)P/PE → S−1B(A,G)/a as φ(a

s+ PE) = a

s+ a for all a

s+ PE ∈

B(G)P/PE. Then from PE = a∩B(G)P it follows that φ is well-defined, and it can

be checked that φ is a ring monomorphism. Hence, B(G)P/PE ≤ S−1B(A,G)/a

is a field extension. Moreover it must be an algebraic field extension because

B(G)P ≤ S−1B(A,G) is an integral ring extension.

Define ψ2 : B(G)P/PE → K as ψ2(

as+PE) = ψ1(

as) for all a

s+PE ∈ B(G)P/P

E.

It is a well-known fact from the field theory that any ring homomorphism from a

field F to an algebraically closed field K can be extended to a ring homomorphism

from F′ to K if F ≤ F′ is an algebraic field extension.

So, there is a ring homomorphism ψ3 : S−1B(A,G)/a → K extending the ring

homomorphism ψ2 : B(G)P/PE → K.

Define ψ4 : S−1B(A,G)→ K as ψ4(xs) = ψ3(

xs

+ a) for all xs∈ S−1B(A,G). It is

clear that ψ4 is a ring homomorphism.

Define ψ5 : B(A,G) → K as ψ5(x) = ψ4(x1) for all x ∈ B(A,G). It can checked

that ψ = ψ5 is a ring homomorphism extending ψ.

�

By the extension procedure given in the last proof we can extend the ring homo-

morphisms SGH : B(G) → Z ≤ C, SGH([S]) = |SH | to ring homomorphisms from

B(A,G) to C. For example, by using the above procedure if we extend SGH we get

SGH,h where h ∈ H is arbitrary, as we already know from 8.1.

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